किसी \(AB_2\) प्रकार के विलेय के (4,g) को (200,g) जल में घोलने पर \(\Delta T_f=0.744,K\) है। यदि वियोजन (50%) है और \(K_f=1.86,K,kg,mol^{-1}\), तो वास्तविक मोलर द्रव्यमान क्या होगा?
When (4,g) of an \(AB_2\)-type solute is dissolved in (200,g) water, \(\Delta T_f=0.744,K\). If dissociation is (50%) and \(K_f=1.86,K,kg,mol^{-1}\), what is the true molar mass?
#freezing point depression
#partial dissociation
#molar mass
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A \(40,g,mol^{-1}\)
B \(50,g,mol^{-1}\)
C \(60,g,mol^{-1}\)
D \(80,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
B. \(50,g,mol^{-1}\)
Step 1
Concept
\(AB_2\) के (50%) वियोजन पर (i=1+2(0.5)=2)। / For (50%) dissociation of \(AB_2\), (i=1+2(0.5)=2).
Step 2
Why this answer is correct
\(\Delta T_f=iK_fm\), इसलिए \(m=\frac{0.744}{2\times1.86}=0.2\)। / From \(\Delta T_f=iK_fm\), \(m=\frac{0.744}{2\times1.86}=0.2\).
Step 3
Exam Tip
(200,g=0.2,kg), मोल \(0.2\times0.2=0.04\), अतः मोलर द्रव्यमान \(=\frac{4}{0.04}=100,g,mol^{-1}\)। / (200,g=0.2,kg), moles \(=0.2\times0.2=0.04\), so molar mass \(=\frac{4}{0.04}=100,g,mol^{-1}\).
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किसी विलेय के (3,g) से (250,mL) विलयन बनाया गया। (300,K) पर परासरण दाब (0.738,atm) है। यदि विलेय का (i=0.75) है, तो वास्तविक मोलर द्रव्यमान क्या होगा?
A (250,mL) solution is prepared from (3,g) solute. At (300,K), osmotic pressure is (0.738,atm). If the solute has (i=0.75), what is the true molar mass?
#osmotic pressure
#association
#true molar mass
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A \(200,g,mol^{-1}\)
B \(300,g,mol^{-1}\)
C \(400,g,mol^{-1}\)
D \(500,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
C. \(400,g,mol^{-1}\)
Step 1
Concept
\(\pi=iCRT\), इसलिए \(C=\frac{0.738}{0.75\times0.082\times300}=0.04,M\)। / From \(\pi=iCRT\), \(C=\frac{0.738}{0.75\times0.082\times300}=0.04,M\).
Step 2
Why this answer is correct
(250,mL=0.25,L), अतः मोल \(0.04\times0.25=0.01\) हैं। / (250,mL=0.25,L), so moles \(=0.04\times0.25=0.01\).
Step 3
Exam Tip
मोलर द्रव्यमान \(=\frac{3}{0.01}=300,g,mol^{-1}\)। / Molar mass \(=\frac{3}{0.01}=300,g,mol^{-1}\).
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वाष्प दाब विधि में विलेय का मोल अंश (0.05) मिला। (2,g) विलेय (18,g) जल में घोला गया था। विलेय का मोलर द्रव्यमान लगभग कितना होगा?
In the vapour pressure method, the mole fraction of solute is found to be (0.05). (2,g) solute was dissolved in (18,g) water. What is the approximate molar mass of the solute?
#vapour pressure method
#mole fraction
#molar mass
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A \(38,g,mol^{-1}\)
B \(40,g,mol^{-1}\)
C \(42,g,mol^{-1}\)
D \(50,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
A. \(38,g,mol^{-1}\)
Step 1
Concept
जल के मोल \(\frac{18}{18}=1\) हैं। / Moles of water \(=\frac{18}{18}=1\).
Step 2
Why this answer is correct
\(0.05=\frac{n_2}{1+n_2}\), इसलिए \(n_2=\frac{0.05}{0.95}=0.0526\) मोल। / \(0.05=\frac{n_2}{1+n_2}\), so \(n_2=\frac{0.05}{0.95}=0.0526\) mol.
Step 3
Exam Tip
मोलर द्रव्यमान \(=\frac{2}{0.0526}\approx38,g,mol^{-1}\)। / Molar mass \(=\frac{2}{0.0526}\approx38,g,mol^{-1}\).
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\(MgCl_2\) का वास्तविक मोलर द्रव्यमान \(95,g,mol^{-1}\) है। यदि अणुसंख्य विधि से प्रेक्षित मोलर द्रव्यमान \(47.5,g,mol^{-1}\) मिले, तो वियोजन की मात्रा कितनी होगी?
The true molar mass of \(MgCl_2\) is \(95,g,mol^{-1}\). If the observed molar mass from a colligative method is \(47.5,g,mol^{-1}\), what is the degree of dissociation?
#magnesium chloride
#degree of dissociation
#observed molar mass
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A (25%)
B (50%)
C (75%)
D (100%)
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Step 1
Concept
\(i=\frac{95}{47.5}=2\)। / \(i=\frac{95}{47.5}=2\).
Step 2
Why this answer is correct
\(MgCl_2\) पूर्ण वियोजन पर (3) कण देता है, इसलिए \(i=1+2\alpha\)। / \(MgCl_2\) gives (3) particles on complete dissociation, so \(i=1+2\alpha\).
Step 3
Exam Tip
\(2=1+2\alpha\), अतः \(\alpha=0.5\), यानी (50%)। / \(2=1+2\alpha\), so \(\alpha=0.5\), or (50%).
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एक विलेय द्विमर बनाता है। वास्तविक मोलर द्रव्यमान \(120,g,mol^{-1}\) है और प्रेक्षित मोलर द्रव्यमान \(150,g,mol^{-1}\) है। संघटन की मात्रा कितनी होगी?
A solute forms dimers. Its true molar mass is \(120,g,mol^{-1}\) and observed molar mass is \(150,g,mol^{-1}\). What is the degree of association?
#dimer
#association
#abnormal molar mass
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A (20%)
B (30%)
C (40%)
D (50%)
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Step 1
Concept
\(i=\frac{120}{150}=0.8\)। / \(i=\frac{120}{150}=0.8\).
Step 2
Why this answer is correct
द्विमर बनने पर \(i=1-\frac{\alpha}{2}\) होता है। / For dimer formation, \(i=1-\frac{\alpha}{2}\).
Step 3
Exam Tip
\(0.8=1-\frac{\alpha}{2}\), इसलिए \(\alpha=0.4\), अर्थात (40%)। / \(0.8=1-\frac{\alpha}{2}\), so \(\alpha=0.4\), meaning (40%).
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एक \(AB_3\) प्रकार का विलेय (40%) वियोजित है। यदि प्रेक्षित मोलर द्रव्यमान \(62.5,g,mol^{-1}\) है, तो वास्तविक मोलर द्रव्यमान क्या होगा?
An \(AB_3\)-type solute is (40%) dissociated. If the observed molar mass is \(62.5,g,mol^{-1}\), what is the true molar mass?
#AB3 electrolyte
#true molar mass
#partial dissociation
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A \(100,g,mol^{-1}\)
B \(125,g,mol^{-1}\)
C \(137.5,g,mol^{-1}\)
D \(150,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
C. \(137.5,g,mol^{-1}\)
Step 1
Concept
\(AB_3\) के लिए \(i=1+3\alpha\)। / For \(AB_3\), \(i=1+3\alpha\).
Step 2
Why this answer is correct
\(\alpha=0.40\), इसलिए (i=1+1.2=2.2)। / With \(\alpha=0.40\), (i=1+1.2=2.2).
Step 3
Exam Tip
वास्तविक मोलर द्रव्यमान \(=2.2\times62.5=137.5,g,mol^{-1}\)। / True molar mass \(=2.2\times62.5=137.5,g,mol^{-1}\).
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किसी (0.04,M) विलयन का (300,K) पर परासरण दाब (1.476,atm) है। यदि वास्तविक मोलर द्रव्यमान \(150,g,mol^{-1}\) है, तो प्रेक्षित मोलर द्रव्यमान कितना होगा?
A (0.04,M) solution has osmotic pressure (1.476,atm) at (300,K). If the true molar mass is \(150,g,mol^{-1}\), what will be the observed molar mass?
#osmotic pressure
#observed molar mass
#van't Hoff factor
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A \(75,g,mol^{-1}\)
B \(90,g,mol^{-1}\)
C \(100,g,mol^{-1}\)
D \(120,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
C. \(100,g,mol^{-1}\)
Step 1
Concept
\(\pi=iCRT\), इसलिए \(i=\frac{1.476}{0.04\times0.082\times300}=1.5\)। / From \(\pi=iCRT\), \(i=\frac{1.476}{0.04\times0.082\times300}=1.5\).
Step 2
Why this answer is correct
\(प्रेक्षित मोलर द्रव्यमान (=\frac{\)वास्तविक मोलर द्रव्यमान}{i})। \(/ Observed molar mass (=\frac{\)true molar mass}{i}).
Step 3
Exam Tip
\(\frac{150}{1.5}=100,g,mol^{-1}\)। / \(\frac{150}{1.5}=100,g,mol^{-1}\).
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एक छात्र ने (125,g) विलायक को (0.125,g) मान लिया। हिमांक विधि से निकला मोलर द्रव्यमान सही मान की तुलना में कैसा होगा?
A student mistakenly takes (125,g) solvent as (0.125,g). In the freezing point method, how will the calculated molar mass compare with the correct value?
#unit error
#freezing point method
#molar mass
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A (1000) गुना अधिक / (1000) times larger
B (1000) गुना कम / (1000) times smaller
C बराबर / Equal
D (125) गुना कम / (125) times smaller
Explanation opens after your attempt
Correct Answer
A. (1000) गुना अधिक / (1000) times larger
Step 1
Concept
(125,g=0.125,kg) सही मान है, लेकिन (0.125,g=0.000125,kg) लिया गया। / Correctly (125,g=0.125,kg), but (0.125,g=0.000125,kg) was used.
Step 2
Why this answer is correct
विलायक का किलोग्राम द्रव्यमान (1000) गुना कम मानने से मोल (1000) गुना कम निकलेंगे। / Taking solvent mass (1000) times smaller gives moles (1000) times smaller.
Step 3
Exam Tip
मोलर द्रव्यमान द्रव्यमान को मोल से भाग देने पर मिलता है, इसलिए वह (1000) गुना अधिक निकलेगा। / Since molar mass is mass divided by moles, it becomes (1000) times larger.
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किसी विलेय के (2.5,g) को (500,g) जल में घोलने पर \(\Delta T_f=0.2325,K\) है। यदि (i=1.25), तो वास्तविक मोलर द्रव्यमान क्या होगा?
When (2.5,g) solute is dissolved in (500,g) water, \(\Delta T_f=0.2325,K\). If (i=1.25), what is the true molar mass?
#freezing point depression
#i correction
#true molar mass
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A \(40,g,mol^{-1}\)
B \(50,g,mol^{-1}\)
C \(60,g,mol^{-1}\)
D \(80,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
B. \(50,g,mol^{-1}\)
Step 1
Concept
प्रभावी मोललता \(\frac{0.2325}{1.86}=0.125\) है। / Effective molality \(=\frac{0.2325}{1.86}=0.125\).
Step 2
Why this answer is correct
वास्तविक मोललता \(\frac{0.125}{1.25}=0.1\) होगी। / True molality \(=\frac{0.125}{1.25}=0.1\).
Step 3
Exam Tip
(500,g=0.5,kg), मोल \(0.1\times0.5=0.05\), अतः मोलर द्रव्यमान \(=\frac{2.5}{0.05}=50,g,mol^{-1}\)। / (500,g=0.5,kg), moles \(=0.1\times0.5=0.05\), so molar mass \(=\frac{2.5}{0.05}=50,g,mol^{-1}\).
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किसी विलेय के (4.8,g) को (400,g) विलायक में घोलने पर \(\Delta T_b=0.156,K\) है। यदि \(K_b=0.52,K,kg,mol^{-1}\) और (i=0.75), तो वास्तविक मोलर द्रव्यमान क्या होगा?
When (4.8,g) solute is dissolved in (400,g) solvent, \(\Delta T_b=0.156,K\). If \(K_b=0.52,K,kg,mol^{-1}\) and (i=0.75), what is the true molar mass?
#boiling point elevation
#association
#true molar mass
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A \(40,g,mol^{-1}\)
B \(48,g,mol^{-1}\)
C \(60,g,mol^{-1}\)
D \(80,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
C. \(60,g,mol^{-1}\)
Step 1
Concept
\(\Delta T_b=iK_bm\), इसलिए \(m=\frac{0.156}{0.75\times0.52}=0.4\)। / From \(\Delta T_b=iK_bm\), \(m=\frac{0.156}{0.75\times0.52}=0.4\).
Step 2
Why this answer is correct
(400,g=0.4,kg), इसलिए मोल \(0.4\times0.4=0.16\) हैं। / (400,g=0.4,kg), so moles \(=0.4\times0.4=0.16\).
Step 3
Exam Tip
मोलर द्रव्यमान \(=\frac{4.8}{0.16}=30,g,mol^{-1}\)। / Molar mass \(=\frac{4.8}{0.16}=30,g,mol^{-1}\).
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किसी विलयन की वास्तविक मोलरता (0.03,M) है और (i=1.8) है। यदि विलेय का वास्तविक मोलर द्रव्यमान \(180,g,mol^{-1}\) है, तो अणुसंख्य विधि से प्रेक्षित मोलर द्रव्यमान कितना होगा?
A solution has actual molarity (0.03,M) and (i=1.8). If the true molar mass of solute is \(180,g,mol^{-1}\), what observed molar mass will be obtained by a colligative method?
#observed molar mass
#van't Hoff factor
#conceptual numerical
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A \(60,g,mol^{-1}\)
B \(90,g,mol^{-1}\)
C \(100,g,mol^{-1}\)
D \(120,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
C. \(100,g,mol^{-1}\)
Step 1
Concept
प्रेक्षित मोलर द्रव्यमान पर सांद्रता का अलग से असर नहीं, (i) का असर मुख्य है। / For observed molar mass, the correction depends mainly on (i), not separately on concentration here.
Step 2
Why this answer is correct
\((M_{\)obs\(}=\frac{M_{\)true}}{i})। \(/ (M_{\)obs\(}=\frac{M_{\)true}}{i}).
Step 3
Exam Tip
\(\frac{180}{1.8}=100,g,mol^{-1}\)। / \(\frac{180}{1.8}=100,g,mol^{-1}\).
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किसी पदार्थ का (60%) द्विमरीकरण होता है। यदि प्रेक्षित मोलर द्रव्यमान \(200,g,mol^{-1}\) है, तो वास्तविक मोलर द्रव्यमान क्या होगा?
A substance undergoes (60%) dimerization. If its observed molar mass is \(200,g,mol^{-1}\), what is the true molar mass?
#dimerization
#true molar mass
#association
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A \(100,g,mol^{-1}\)
B \(120,g,mol^{-1}\)
C \(140,g,mol^{-1}\)
D \(160,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
C. \(140,g,mol^{-1}\)
Step 1
Concept
द्विमरीकरण के लिए \(i=1-\frac{\alpha}{2}=1-\frac{0.6}{2}=0.7\)। / For dimerization, \(i=1-\frac{\alpha}{2}=1-\frac{0.6}{2}=0.7\).
Step 2
Why this answer is correct
वास्तविक मोलर द्रव्यमान \(=i\times\) प्रेक्षित मोलर द्रव्यमान। / True molar mass \(=i\times\) observed molar mass.
Step 3
Exam Tip
\(0.7\times200=140,g,mol^{-1}\)। / \(0.7\times200=140,g,mol^{-1}\).
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एक पदार्थ (45%) त्रिमर बनाता है। यदि वास्तविक मोलर द्रव्यमान \(210,g,mol^{-1}\) है, तो प्रेक्षित मोलर द्रव्यमान लगभग कितना होगा?
A substance forms trimers to the extent of (45%). If its true molar mass is \(210,g,mol^{-1}\), what will be the approximate observed molar mass?
#trimer association
#observed molar mass
#expert
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A \(250,g,mol^{-1}\)
B \(280,g,mol^{-1}\)
C \(300,g,mol^{-1}\)
D \(350,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
C. \(300,g,mol^{-1}\)
Step 1
Concept
त्रिमर संघटन के लिए \(i=1-\frac{2\alpha}{3}\)। / For trimer association, \(i=1-\frac{2\alpha}{3}\).
Step 2
Why this answer is correct
\(\alpha=0.45\), इसलिए \(i=1-\frac{0.90}{3}=0.70\)। / With \(\alpha=0.45\), \(i=1-\frac{0.90}{3}=0.70\).
Step 3
Exam Tip
प्रेक्षित मोलर द्रव्यमान \(=\frac{210}{0.70}=300,g,mol^{-1}\)। / Observed molar mass \(=\frac{210}{0.70}=300,g,mol^{-1}\).
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एक (AB) प्रकार का विलेय (60%) वियोजित है। यदि प्रेक्षित मोलर द्रव्यमान \(75,g,mol^{-1}\) है, तो वास्तविक मोलर द्रव्यमान कितना होगा?
An (AB)-type solute is (60%) dissociated. If its observed molar mass is \(75,g,mol^{-1}\), what is the true molar mass?
#AB electrolyte
#true molar mass
#partial dissociation
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A \(90,g,mol^{-1}\)
B \(105,g,mol^{-1}\)
C \(120,g,mol^{-1}\)
D \(150,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
C. \(120,g,mol^{-1}\)
Step 1
Concept
(AB) के लिए \(i=1+\alpha=1+0.6=1.6\)। / For (AB), \(i=1+\alpha=1+0.6=1.6\).
Step 2
Why this answer is correct
\(वास्तविक मोलर द्रव्यमान (=i\times M_{\)obs})। \(/ True molar mass (=i\times M_{\)obs}).
Step 3
Exam Tip
\(1.6\times75=120,g,mol^{-1}\)। / \(1.6\times75=120,g,mol^{-1}\).
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किसी \(A_2B\) प्रकार के विलेय का प्रेक्षित मोलर द्रव्यमान वास्तविक का \(\frac{1}{2.4}\) है। वियोजन की मात्रा कितनी होगी?
The observed molar mass of an \(A_2B\)-type solute is \(\frac{1}{2.4}\) of the true value. What is the degree of dissociation?
#A2B electrolyte
#degree of dissociation
#observed molar mass
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A (50%)
B (60%)
C (70%)
D (80%)
Explanation opens after your attempt
Step 1
Concept
\((M_{\)obs\(}=\frac{M_{\)true}}{2.4}), इसलिए (i=2.4)। \(/ Since (M_{\)obs\(}=\frac{M_{\)true\(}}{2.4}), (i=2.4).\)
Step 2
Why this answer is correct
\(A_2B\) पूर्ण वियोजन पर (3) कण देता है, अतः \(i=1+2\alpha\)। / \(A_2B\) gives (3) particles on complete dissociation, so \(i=1+2\alpha\).
Step 3
Exam Tip
\(2.4=1+2\alpha\), इसलिए \(\alpha=0.7\), यानी (70%)। / \(2.4=1+2\alpha\), so \(\alpha=0.7\), or (70%).
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अवाष्पशील विलेय के (3,g) को (72,g) जल में घोला गया। वाष्प दाब में आपेक्षिक कमी (0.04) है। विलेय का मोलर द्रव्यमान लगभग कितना होगा?
(3,g) of a non-volatile solute is dissolved in (72,g) water. The relative lowering of vapour pressure is (0.04). What is the approximate molar mass of the solute?
#relative lowering vapour pressure
#mole fraction
#molar mass
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A \(18,g,mol^{-1}\)
B \(20,g,mol^{-1}\)
C \(36,g,mol^{-1}\)
D \(48,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
A. \(18,g,mol^{-1}\)
Step 1
Concept
जल के मोल \(\frac{72}{18}=4\) और \(x_2=0.04\) है। / Moles of water \(=\frac{72}{18}=4\), and \(x_2=0.04\).
Step 2
Why this answer is correct
\(0.04=\frac{n_2}{4+n_2}\), इसलिए \(n_2=\frac{0.16}{0.96}=0.1667\) मोल। / \(0.04=\frac{n_2}{4+n_2}\), so \(n_2=\frac{0.16}{0.96}=0.1667\) mol.
Step 3
Exam Tip
मोलर द्रव्यमान \(=\frac{3}{0.1667}\approx18,g,mol^{-1}\)। / Molar mass \(=\frac{3}{0.1667}\approx18,g,mol^{-1}\).
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एक (AB) विलेय के (3.6,g) को (300,g) जल में घोलने पर \(\Delta T_f=0.558,K\) है। यदि वियोजन (25%) है, तो वास्तविक मोलर द्रव्यमान क्या होगा?
When (3.6,g) of an (AB) solute is dissolved in (300,g) water, \(\Delta T_f=0.558,K\). If dissociation is (25%), what is the true molar mass?
#freezing point depression
#partial dissociation
#true molar mass
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A \(40,g,mol^{-1}\)
B \(50,g,mol^{-1}\)
C \(60,g,mol^{-1}\)
D \(80,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
B. \(50,g,mol^{-1}\)
Step 1
Concept
(AB) के (25%) वियोजन पर (i=1.25)। / For (25%) dissociation of (AB), (i=1.25).
Step 2
Why this answer is correct
वास्तविक मोललता \(m=\frac{0.558}{1.25\times1.86}=0.24\)। / True molality \(m=\frac{0.558}{1.25\times1.86}=0.24\).
Step 3
Exam Tip
(300,g=0.3,kg), मोल \(0.24\times0.3=0.072\), इसलिए मोलर द्रव्यमान \(=\frac{3.6}{0.072}=50,g,mol^{-1}\)। / (300,g=0.3,kg), moles \(=0.24\times0.3=0.072\), so molar mass \(=\frac{3.6}{0.072}=50,g,mol^{-1}\).
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किसी विलेय के (2,g) से (500,mL) विलयन बनाया गया। (300,K) पर \(\pi=0.615,atm\) है। यदि (i=1.25), तो वास्तविक मोलर द्रव्यमान कितना होगा?
A (500,mL) solution is prepared from (2,g) solute. At (300,K), \(\pi=0.615,atm\). If (i=1.25), what is the true molar mass?
#osmotic pressure
#i correction
#true molar mass
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A \(120,g,mol^{-1}\)
B \(160,g,mol^{-1}\)
C \(200,g,mol^{-1}\)
D \(250,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
B. \(160,g,mol^{-1}\)
Step 1
Concept
\(C=\frac{0.615}{1.25\times0.082\times300}=0.02,M\)। / \(C=\frac{0.615}{1.25\times0.082\times300}=0.02,M\).
Step 2
Why this answer is correct
(500,mL=0.5,L), इसलिए मोल \(0.02\times0.5=0.01\) हैं। / (500,mL=0.5,L), so moles \(=0.02\times0.5=0.01\).
Step 3
Exam Tip
मोलर द्रव्यमान \(=\frac{2}{0.01}=200,g,mol^{-1}\)। / Molar mass \(=\frac{2}{0.01}=200,g,mol^{-1}\).
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यदि किसी विलेय का प्रेक्षित मोलर द्रव्यमान वास्तविक का (0.4) गुना है, तो (i) और संभावित व्यवहार क्या होगा?
If the observed molar mass of a solute is (0.4) times the true value, what are (i) and the probable behaviour?
#observed molar mass
#ratio
#dissociation
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A (i=0.4), संघटन / (i=0.4), association
B (i=1.4), वियोजन / (i=1.4), dissociation
C (i=2.5), वियोजन / (i=2.5), dissociation
D (i=2.5), संघटन / (i=2.5), association
Explanation opens after your attempt
Correct Answer
C. (i=2.5), वियोजन / (i=2.5), dissociation
Step 1
Concept
\((M_{\)obs\(}=0.4M_{\)true})। \(/ (M_{\)obs\(}=0.4M_{\)true}).
Step 2
Why this answer is correct
\((i=\frac{M_{\)true\(}}{M_{\)obs}}=\frac{1}{0.4}=2.5)। \(/ (i=\frac{M_{\)true\(}}{M_{\)obs\(}}=\frac{1}{0.4}=2.5).\)
Step 3
Exam Tip
(i>1) बताता है कि विलेय वियोजित होकर अधिक कण बना रहा है। / (i>1) shows dissociation into more particles.
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यदि किसी विलेय का प्रेक्षित मोलर द्रव्यमान वास्तविक का (1.6) गुना है, तो (i) और संभावित व्यवहार क्या होगा?
If the observed molar mass of a solute is (1.6) times the true value, what are (i) and the probable behaviour?
#association
#observed molar mass
#ratio
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A (i=0.625), संघटन / (i=0.625), association
B (i=1.6), वियोजन / (i=1.6), dissociation
C (i=0.8), वियोजन / (i=0.8), dissociation
D (i=2.6), संघटन / (i=2.6), association
Explanation opens after your attempt
Correct Answer
A. (i=0.625), संघटन / (i=0.625), association
Step 1
Concept
\((M_{\)obs\(}=1.6M_{\)true})। \(/ (M_{\)obs\(}=1.6M_{\)true}).
Step 2
Why this answer is correct
\(i=\frac{1}{1.6}=0.625\)। / \(i=\frac{1}{1.6}=0.625\).
Step 3
Exam Tip
(i<1) होने का अर्थ है स्वतंत्र कण कम हुए, इसलिए संघटन की संभावना है।
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किसी विलेय के (2.4,g) को (200,g) विलायक में घोलने पर \(\Delta T_b=0.156,K\) है। यदि \(K_b=0.52,K,kg,mol^{-1}\) और (i=1.5), तो वास्तविक मोलर द्रव्यमान क्या होगा?
When (2.4,g) solute is dissolved in (200,g) solvent, \(\Delta T_b=0.156,K\). If \(K_b=0.52,K,kg,mol^{-1}\) and (i=1.5), what is the true molar mass?
#boiling point elevation
#electrolyte
#true molar mass
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A \(40,g,mol^{-1}\)
B \(50,g,mol^{-1}\)
C \(60,g,mol^{-1}\)
D \(80,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
C. \(60,g,mol^{-1}\)
Step 1
Concept
\(m=\frac{0.156}{1.5\times0.52}=0.2\)। / \(m=\frac{0.156}{1.5\times0.52}=0.2\).
Step 2
Why this answer is correct
(200,g=0.2,kg), इसलिए मोल \(0.2\times0.2=0.04\) हैं। / (200,g=0.2,kg), so moles \(=0.2\times0.2=0.04\).
Step 3
Exam Tip
मोलर द्रव्यमान \(=\frac{2.4}{0.04}=60,g,mol^{-1}\)। / Molar mass \(=\frac{2.4}{0.04}=60,g,mol^{-1}\).
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किसी विलेय के (1.8,g) को (150,g) जल में घोलने पर \(\Delta T_f=0.279,K\) है। यदि (i=0.75), तो वास्तविक मोलर द्रव्यमान कितना होगा?
When (1.8,g) solute is dissolved in (150,g) water, \(\Delta T_f=0.279,K\). If (i=0.75), what is the true molar mass?
#association
#freezing point depression
#true molar mass
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A \(45,g,mol^{-1}\)
B \(60,g,mol^{-1}\)
C \(75,g,mol^{-1}\)
D \(90,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
B. \(60,g,mol^{-1}\)
Step 1
Concept
प्रभावी मोललता \(\frac{0.279}{1.86}=0.15\) है। / Effective molality \(=\frac{0.279}{1.86}=0.15\).
Step 2
Why this answer is correct
वास्तविक मोललता \(\frac{0.15}{0.75}=0.2\) होगी। / True molality \(=\frac{0.15}{0.75}=0.2\).
Step 3
Exam Tip
(150,g=0.15,kg), मोल \(0.2\times0.15=0.03\), इसलिए मोलर द्रव्यमान \(=\frac{1.8}{0.03}=60,g,mol^{-1}\)। / (150,g=0.15,kg), moles \(=0.2\times0.15=0.03\), so molar mass \(=\frac{1.8}{0.03}=60,g,mol^{-1}\).
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किसी \(A_2B_3\) प्रकार के विलेय का (i=3.4) है। पूर्ण वियोजन पर (5) कण बनते हैं। वियोजन की मात्रा कितनी होगी?
An \(A_2B_3\)-type solute has (i=3.4). Complete dissociation gives (5) particles. What is the degree of dissociation?
#A2B3 electrolyte
#degree of dissociation
#van't Hoff factor
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A (50%)
B (60%)
C (70%)
D (80%)
Explanation opens after your attempt
Step 1
Concept
\(A_2B_3\) पूर्ण वियोजन पर (5) कण देता है। / \(A_2B_3\) gives (5) particles on complete dissociation.
Step 2
Why this answer is correct
(i=1+\alpha(5-1)=1+4\alpha)। / (i=1+\alpha(5-1)=1+4\alpha).
Step 3
Exam Tip
\(3.4=1+4\alpha\), इसलिए \(\alpha=0.6\), अर्थात (60%)। / \(3.4=1+4\alpha\), so \(\alpha=0.6\), meaning (60%).
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\(FeCl_3\) का (60%) वियोजन है। यदि प्रेक्षित मोलर द्रव्यमान \(81.25,g,mol^{-1}\) है, तो वास्तविक मोलर द्रव्यमान कितना होगा?
\(FeCl_3\) is (60%) dissociated. If its observed molar mass is \(81.25,g,mol^{-1}\), what is the true molar mass?
#ferric chloride
#true molar mass
#partial dissociation
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A \(130,g,mol^{-1}\)
B \(162.5,g,mol^{-1}\)
C \(195,g,mol^{-1}\)
D \(243.75,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
C. \(195,g,mol^{-1}\)
Step 1
Concept
\(FeCl_3\) पूर्ण वियोजन पर (4) कण देता है, इसलिए \(i=1+3\alpha\)। / \(FeCl_3\) gives (4) particles on complete dissociation, so \(i=1+3\alpha\).
Step 2
Why this answer is correct
\(\alpha=0.60\), अतः (i=1+1.8=2.8)। / With \(\alpha=0.60\), (i=1+1.8=2.8).
Step 3
Exam Tip
वास्तविक मोलर द्रव्यमान \(=2.8\times81.25=227.5,g,mol^{-1}\)। / True molar mass \(=2.8\times81.25=227.5,g,mol^{-1}\).
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वाष्प दाब विधि में \(p^0=120,mm\) और (p=114,mm) है। (3,g) विलेय (54,g) जल में घुला है। विलेय का मोलर द्रव्यमान लगभग कितना होगा?
In vapour pressure method, \(p^0=120,mm\) and (p=114,mm). (3,g) solute is dissolved in (54,g) water. What is the approximate molar mass of solute?
#vapour pressure lowering
#Raoult law
#molar mass
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A \(18,g,mol^{-1}\)
B \(19,g,mol^{-1}\)
C \(20,g,mol^{-1}\)
D \(24,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
B. \(19,g,mol^{-1}\)
Step 1
Concept
आपेक्षिक कमी \(\frac{120-114}{120}=0.05\), इसलिए \(x_2=0.05\)। / Relative lowering \(=\frac{120-114}{120}=0.05\), so \(x_2=0.05\).
Step 2
Why this answer is correct
जल के मोल \(\frac{54}{18}=3\), अतः \(0.05=\frac{n_2}{3+n_2}\) से \(n_2=0.1579\)। / Moles of water \(=\frac{54}{18}=3\), so \(0.05=\frac{n_2}{3+n_2}\) gives \(n_2=0.1579\).
Step 3
Exam Tip
मोलर द्रव्यमान \(=\frac{3}{0.1579}\approx19,g,mol^{-1}\)। / Molar mass \(=\frac{3}{0.1579}\approx19,g,mol^{-1}\).
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एक विलेय का प्रेक्षित मोलर द्रव्यमान \(72,g,mol^{-1}\) है। यदि वह \(A_2B\) प्रकार का है और (40%) वियोजित है, तो वास्तविक मोलर द्रव्यमान क्या होगा?
A solute has observed molar mass \(72,g,mol^{-1}\). If it is of \(A_2B\)-type and (40%) dissociated, what is the true molar mass?
#A2B electrolyte
#true molar mass
#partial dissociation
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A \(100.8,g,mol^{-1}\)
B \(115.2,g,mol^{-1}\)
C \(129.6,g,mol^{-1}\)
D \(144,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
C. \(129.6,g,mol^{-1}\)
Step 1
Concept
\(A_2B\) के लिए \(i=1+2\alpha\)। / For \(A_2B\), \(i=1+2\alpha\).
Step 2
Why this answer is correct
\(\alpha=0.40\), इसलिए (i=1+0.8=1.8)। / With \(\alpha=0.40\), (i=1+0.8=1.8).
Step 3
Exam Tip
वास्तविक मोलर द्रव्यमान \(=1.8\times72=129.6,g,mol^{-1}\)। / True molar mass \(=1.8\times72=129.6,g,mol^{-1}\).
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यदि किसी विलेय का वास्तविक मोलर द्रव्यमान \(180,g,mol^{-1}\) है और (30%) द्विमरीकरण होता है, तो प्रेक्षित मोलर द्रव्यमान लगभग कितना होगा?
If the true molar mass of a solute is \(180,g,mol^{-1}\) and (30%) dimerization occurs, what will be the approximate observed molar mass?
#dimerization
#observed molar mass
#association
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A \(180,g,mol^{-1}\)
B \(200,g,mol^{-1}\)
C \(211.8,g,mol^{-1}\)
D \(240,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
C. \(211.8,g,mol^{-1}\)
Step 1
Concept
द्विमरीकरण के लिए \(i=1-\frac{\alpha}{2}\)। / For dimerization, \(i=1-\frac{\alpha}{2}\).
Step 2
Why this answer is correct
\(\alpha=0.30\), इसलिए (i=0.85)। / With \(\alpha=0.30\), (i=0.85).
Step 3
Exam Tip
प्रेक्षित मोलर द्रव्यमान \(=\frac{180}{0.85}\approx211.8,g,mol^{-1}\)। / Observed molar mass \(=\frac{180}{0.85}\approx211.8,g,mol^{-1}\).
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एक विलयन में (2.4,g) विलेय (400,mL) में है। (300,K) पर परासरण दाब (0.492,atm) है। यदि (i=0.8), तो वास्तविक मोलर द्रव्यमान क्या होगा?
A solution contains (2.4,g) solute in (400,mL). Its osmotic pressure at (300,K) is (0.492,atm). If (i=0.8), what is the true molar mass?
#osmotic pressure
#association
#true molar mass
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A \(200,g,mol^{-1}\)
B \(240,g,mol^{-1}\)
C \(300,g,mol^{-1}\)
D \(360,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
B. \(240,g,mol^{-1}\)
Step 1
Concept
\(C=\frac{0.492}{0.8\times0.082\times300}=0.025,M\)। / \(C=\frac{0.492}{0.8\times0.082\times300}=0.025,M\).
Step 2
Why this answer is correct
(400,mL=0.4,L), इसलिए मोल \(0.025\times0.4=0.01\) हैं। / (400,mL=0.4,L), so moles \(=0.025\times0.4=0.01\).
Step 3
Exam Tip
मोलर द्रव्यमान \(=\frac{2.4}{0.01}=240,g,mol^{-1}\)। / Molar mass \(=\frac{2.4}{0.01}=240,g,mol^{-1}\).
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किसी विलेय के (3,g) को (250,g) जल में घोलने पर \(\Delta T_f=0.465,K\) है। यदि (i=1.5), तो वास्तविक मोलर द्रव्यमान कितना होगा?
When (3,g) solute is dissolved in (250,g) water, \(\Delta T_f=0.465,K\). If (i=1.5), what is the true molar mass?
#freezing point depression
#i factor
#molar mass
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A \(60,g,mol^{-1}\)
B \(72,g,mol^{-1}\)
C \(80,g,mol^{-1}\)
D \(90,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
B. \(72,g,mol^{-1}\)
Step 1
Concept
प्रभावी मोललता \(\frac{0.465}{1.86}=0.25\) है। / Effective molality \(=\frac{0.465}{1.86}=0.25\).
Step 2
Why this answer is correct
वास्तविक मोललता \(\frac{0.25}{1.5}=0.1667\) होगी। / True molality \(=\frac{0.25}{1.5}=0.1667\).
Step 3
Exam Tip
(250,g=0.25,kg), मोल \(0.1667\times0.25=0.0417\), अतः मोलर द्रव्यमान \(\frac{3}{0.0417}\approx72,g,mol^{-1}\)। / (250,g=0.25,kg), moles \(=0.1667\times0.25=0.0417\), so molar mass \(=\frac{3}{0.0417}\approx72,g,mol^{-1}\).
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किसी विलेय के (5,g) को (500,g) विलायक में घोलने पर \(\Delta T_b=0.078,K\) है। \(K_b=0.52,K,kg,mol^{-1}\) और (i=1.5) हो तो वास्तविक मोलर द्रव्यमान क्या होगा?
When (5,g) solute is dissolved in (500,g) solvent, \(\Delta T_b=0.078,K\). If \(K_b=0.52,K,kg,mol^{-1}\) and (i=1.5), what is the true molar mass?
#boiling point elevation
#van't Hoff factor
#true molar mass
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A \(100,g,mol^{-1}\)
B \(125,g,mol^{-1}\)
C \(150,g,mol^{-1}\)
D \(200,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
A. \(100,g,mol^{-1}\)
Step 1
Concept
वास्तविक मोललता \(m=\frac{0.078}{1.5\times0.52}=0.1\) है। / True molality \(m=\frac{0.078}{1.5\times0.52}=0.1\).
Step 2
Why this answer is correct
(500,g=0.5,kg), इसलिए मोल \(0.1\times0.5=0.05\) हैं। / (500,g=0.5,kg), so moles \(=0.1\times0.5=0.05\).
Step 3
Exam Tip
मोलर द्रव्यमान \(=\frac{5}{0.05}=100,g,mol^{-1}\)। / Molar mass \(=\frac{5}{0.05}=100,g,mol^{-1}\).
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यदि (0.3,m) विलयन में (i=0.6) है, तो अणुसंख्य गुण से बिना (i) सुधार के मोललता कैसी दिखाई देगी?
If a (0.3,m) solution has (i=0.6), what molality will appear from a colligative property without (i) correction?
#effective molality
#association
#molar mass error
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A (0.18,m)
B (0.30,m)
C (0.50,m)
D (0.90,m)
Explanation opens after your attempt
Correct Answer
A. (0.18,m)
Step 1
Concept
अणुसंख्य प्रभाव \(i\times m\) पर निर्भर करता है। / Colligative effect depends on \(i\times m\).
Step 2
Why this answer is correct
\(i\times m=0.6\times0.3=0.18,m\)। / \(i\times m=0.6\times0.3=0.18,m\).
Step 3
Exam Tip
बिना (i) सुधार के यही कम मोललता मानी जाएगी। / Without (i) correction, this lower molality will be assumed.
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किसी विलेय के (2,g) को (0.4,kg) विलायक में घोलने पर प्रभावी मोललता (0.10,m) मिली। यदि (i=0.5), तो वास्तविक मोलर द्रव्यमान क्या होगा?
When (2,g) solute is dissolved in (0.4,kg) solvent, effective molality is found to be (0.10,m). If (i=0.5), what is the true molar mass?
#effective molality
#association
#true molar mass
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A \(20,g,mol^{-1}\)
B \(25,g,mol^{-1}\)
C \(40,g,mol^{-1}\)
D \(50,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
B. \(25,g,mol^{-1}\)
Step 1
Concept
वास्तविक मोललता \(\frac{0.10}{0.5}=0.20,m\) है। / True molality \(=\frac{0.10}{0.5}=0.20,m\).
Step 2
Why this answer is correct
मोल \(0.20\times0.4=0.08\) होंगे। / Moles \(=0.20\times0.4=0.08\).
Step 3
Exam Tip
मोलर द्रव्यमान \(=\frac{2}{0.08}=25,g,mol^{-1}\)। / Molar mass \(=\frac{2}{0.08}=25,g,mol^{-1}\).
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यदि किसी विलेय का वास्तविक मोलर द्रव्यमान \(240,g,mol^{-1}\) है और अणुसंख्य विधि से \(160,g,mol^{-1}\) मिलता है, तो (i) और संभावित व्यवहार क्या होगा?
If the true molar mass of a solute is \(240,g,mol^{-1}\) and colligative method gives \(160,g,mol^{-1}\), what are (i) and the probable behaviour?
#abnormal molar mass
#dissociation
#van't Hoff factor
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A (i=0.67), संघटन / (i=0.67), association
B (i=1.5), वियोजन / (i=1.5), dissociation
C (i=2.4), वियोजन / (i=2.4), dissociation
D (i=1), सामान्य / (i=1), normal
Explanation opens after your attempt
Correct Answer
B. (i=1.5), वियोजन / (i=1.5), dissociation
Step 1
Concept
\((i=\frac{M_{\)true\(}}{M_{\)obs}}=\frac{240}{160}=1.5)। \(/ (i=\frac{M_{\)true\(}}{M_{\)obs\(}}=\frac{240}{160}=1.5).\)
Step 2
Why this answer is correct
(i>1) बताता है कि प्रभावी कणों की संख्या बढ़ी है। / (i>1) means the effective particle number has increased.
Step 3
Exam Tip
यह वियोजन की संभावना दिखाता है। / This suggests dissociation.
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किसी विलेय के (1.5,g) को (100,g) जल में घोलने पर \(\Delta T_f=0.186,K\) है। यदि (i=0.75), तो वास्तविक मोलर द्रव्यमान क्या होगा?
When (1.5,g) solute is dissolved in (100,g) water, \(\Delta T_f=0.186,K\). If (i=0.75), what is the true molar mass?
#association
#freezing point depression
#true molar mass
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A \(50,g,mol^{-1}\)
B \(75,g,mol^{-1}\)
C \(100,g,mol^{-1}\)
D \(150,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
A. \(50,g,mol^{-1}\)
Step 1
Concept
प्रभावी मोललता \(\frac{0.186}{1.86}=0.1\) है। / Effective molality \(=\frac{0.186}{1.86}=0.1\).
Step 2
Why this answer is correct
वास्तविक मोललता \(\frac{0.1}{0.75}=0.1333\) होगी। / True molality \(=\frac{0.1}{0.75}=0.1333\).
Step 3
Exam Tip
(100,g=0.1,kg), मोल (0.01333), इसलिए मोलर द्रव्यमान \(\frac{1.5}{0.01333}\approx112.5,g,mol^{-1}\)। / (100,g=0.1,kg), moles (=0.01333), so molar mass \(=\frac{1.5}{0.01333}\approx112.5,g,mol^{-1}\).
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एक विलेय के (1.8,g) से (300,mL) विलयन बना। (300,K) पर \(\pi=0.492,atm\) है। यदि (i=1.2), तो वास्तविक मोलर द्रव्यमान कितना होगा?
A (300,mL) solution is prepared from (1.8,g) solute. At (300,K), \(\pi=0.492,atm\). If (i=1.2), what is the true molar mass?
#osmotic pressure
#van't Hoff factor
#true molar mass
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A \(90,g,mol^{-1}\)
B \(120,g,mol^{-1}\)
C \(150,g,mol^{-1}\)
D \(180,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
C. \(150,g,mol^{-1}\)
Step 1
Concept
\(C=\frac{0.492}{1.2\times0.082\times300}=0.0167,M\)। / \(C=\frac{0.492}{1.2\times0.082\times300}=0.0167,M\).
Step 2
Why this answer is correct
(300,mL=0.3,L), इसलिए मोल \(0.0167\times0.3=0.005\) हैं। / (300,mL=0.3,L), so moles \(=0.0167\times0.3=0.005\).
Step 3
Exam Tip
मोलर द्रव्यमान \(=\frac{1.8}{0.005}=360,g,mol^{-1}\)। / Molar mass \(=\frac{1.8}{0.005}=360,g,mol^{-1}\).
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यदि \(K_f=1.86,K,kg,mol^{-1}\), विलेय (4,g), विलायक (200,g), मोलर द्रव्यमान \(100,g,mol^{-1}\), और (i=1.5) है, तो \(\Delta T_f\) कितना होगा?
If \(K_f=1.86,K,kg,mol^{-1}\), solute mass is (4,g), solvent mass is (200,g), molar mass is \(100,g,mol^{-1}\), and (i=1.5), what is \(\Delta T_f\)?
#reverse calculation
#freezing point depression
#i factor
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A (0.279,K)
B (0.372,K)
C (0.465,K)
D (0.558,K)
Explanation opens after your attempt
Correct Answer
D. (0.558,K)
Step 1
Concept
विलेय के मोल \(\frac{4}{100}=0.04\) हैं। / Moles of solute \(=\frac{4}{100}=0.04\).
Step 2
Why this answer is correct
(200,g=0.2,kg), इसलिए मोललता (0.2,m) है। / (200,g=0.2,kg), so molality is (0.2,m).
Step 3
Exam Tip
\(\Delta T_f=iK_fm=1.5\times1.86\times0.2=0.558,K\)। / \(\Delta T_f=iK_fm=1.5\times1.86\times0.2=0.558,K\).
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यदि \(K_b=0.52,K,kg,mol^{-1}\), विलेय (3,g), विलायक (250,g), मोलर द्रव्यमान \(60,g,mol^{-1}\), और (i=2) है, तो \(\Delta T_b\) कितना होगा?
If \(K_b=0.52,K,kg,mol^{-1}\), solute mass is (3,g), solvent mass is (250,g), molar mass is \(60,g,mol^{-1}\), and (i=2), what is \(\Delta T_b\)?
#boiling point elevation
#reverse numerical
#molar mass
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A (0.104,K)
B (0.208,K)
C (0.312,K)
D (0.416,K)
Explanation opens after your attempt
Correct Answer
B. (0.208,K)
Step 1
Concept
विलेय के मोल \(\frac{3}{60}=0.05\) हैं। / Moles of solute \(=\frac{3}{60}=0.05\).
Step 2
Why this answer is correct
(250,g=0.25,kg), इसलिए मोललता (0.2,m) है। / (250,g=0.25,kg), so molality is (0.2,m).
Step 3
Exam Tip
\(\Delta T_b=2\times0.52\times0.2=0.208,K\)। / \(\Delta T_b=2\times0.52\times0.2=0.208,K\).
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किसी विलेय के (4.5,g) को (300,g) जल में घोलने पर \(\Delta T_f=0.837,K\) है। यदि (i=1.5), तो वास्तविक मोलर द्रव्यमान क्या होगा?
When (4.5,g) solute is dissolved in (300,g) water, \(\Delta T_f=0.837,K\). If (i=1.5), what is the true molar mass?
#freezing point depression
#i correction
#molar mass
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A \(50,g,mol^{-1}\)
B \(60,g,mol^{-1}\)
C \(75,g,mol^{-1}\)
D \(90,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
B. \(60,g,mol^{-1}\)
Step 1
Concept
वास्तविक मोललता \(m=\frac{0.837}{1.5\times1.86}=0.3\) है। / True molality \(m=\frac{0.837}{1.5\times1.86}=0.3\).
Step 2
Why this answer is correct
(300,g=0.3,kg), इसलिए मोल \(0.3\times0.3=0.09\) हैं। / (300,g=0.3,kg), so moles \(=0.3\times0.3=0.09\).
Step 3
Exam Tip
मोलर द्रव्यमान \(=\frac{4.5}{0.09}=50,g,mol^{-1}\)। / Molar mass \(=\frac{4.5}{0.09}=50,g,mol^{-1}\).
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अवाष्पशील विलेय के (4,g) को (36,g) जल में घोलने पर वाष्प दाब में आपेक्षिक कमी (0.08) है। विलेय का मोलर द्रव्यमान लगभग कितना होगा?
When (4,g) of a non-volatile solute is dissolved in (36,g) water, the relative lowering of vapour pressure is (0.08). What is the approximate molar mass of the solute?
#vapour pressure lowering
#mole fraction
#molar mass
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A \(23,g,mol^{-1}\)
B \(25,g,mol^{-1}\)
C \(46,g,mol^{-1}\)
D \(50,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
A. \(23,g,mol^{-1}\)
Step 1
Concept
जल के मोल \(\frac{36}{18}=2\) हैं और \(x_2=0.08\)। / Moles of water \(=\frac{36}{18}=2\), and \(x_2=0.08\).
Step 2
Why this answer is correct
\(0.08=\frac{n_2}{2+n_2}\), इसलिए \(n_2=\frac{0.16}{0.92}\approx0.174\) मोल। / \(0.08=\frac{n_2}{2+n_2}\), so \(n_2=\frac{0.16}{0.92}\approx0.174\) mol.
Step 3
Exam Tip
मोलर द्रव्यमान \(=\frac{4}{0.174}\approx23,g,mol^{-1}\)। / Molar mass \(=\frac{4}{0.174}\approx23,g,mol^{-1}\).
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किसी विलेय का (i=2.25) है और प्रेक्षित मोलर द्रव्यमान \(64,g,mol^{-1}\) है। वास्तविक मोलर द्रव्यमान क्या होगा?
A solute has (i=2.25) and observed molar mass \(64,g,mol^{-1}\). What is the true molar mass?
#true molar mass
#observed molar mass
#van't Hoff factor
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A \(128,g,mol^{-1}\)
B \(144,g,mol^{-1}\)
C \(160,g,mol^{-1}\)
D \(192,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
B. \(144,g,mol^{-1}\)
Step 1
Concept
\((M_{\)obs\(}=\frac{M_{\)true}}{i}) होता है। \(/ (M_{\)obs\(}=\frac{M_{\)true}}{i}).
Step 2
Why this answer is correct
\(इसलिए (M_{\)true\(}=iM_{\)obs})। \(/ Therefore (M_{\)true\(}=iM_{\)obs}).
Step 3
Exam Tip
\(2.25\times64=144,g,mol^{-1}\)। / \(2.25\times64=144,g,mol^{-1}\).
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किसी \(AB_2\) विलेय का वास्तविक मोलर द्रव्यमान \(180,g,mol^{-1}\) है और वियोजन (70%) है। प्रेक्षित मोलर द्रव्यमान लगभग कितना होगा?
An \(AB_2\) solute has true molar mass \(180,g,mol^{-1}\) and is (70%) dissociated. What will be the approximate observed molar mass?
#AB2 electrolyte
#observed molar mass
#partial dissociation
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A \(60,g,mol^{-1}\)
B \(75,g,mol^{-1}\)
C \(90,g,mol^{-1}\)
D \(105.9,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
B. \(75,g,mol^{-1}\)
Step 1
Concept
\(AB_2\) के लिए \(i=1+2\alpha\)। / For \(AB_2\), \(i=1+2\alpha\).
Step 2
Why this answer is correct
\(\alpha=0.70\), इसलिए (i=2.4)। / With \(\alpha=0.70\), (i=2.4).
Step 3
Exam Tip
प्रेक्षित मोलर द्रव्यमान \(=\frac{180}{2.4}=75,g,mol^{-1}\)। / Observed molar mass \(=\frac{180}{2.4}=75,g,mol^{-1}\).
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एक विलेय (60%) त्रिमर बनाता है। यदि वास्तविक मोलर द्रव्यमान \(180,g,mol^{-1}\) है, तो प्रेक्षित मोलर द्रव्यमान कितना होगा?
A solute forms trimers to the extent of (60%). If its true molar mass is \(180,g,mol^{-1}\), what will be the observed molar mass?
#trimer association
#observed molar mass
#expert numerical
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A \(240,g,mol^{-1}\)
B \(270,g,mol^{-1}\)
C \(300,g,mol^{-1}\)
D \(360,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
C. \(300,g,mol^{-1}\)
Step 1
Concept
त्रिमर संघटन के लिए \(i=1-\frac{2\alpha}{3}\)। / For trimer association, \(i=1-\frac{2\alpha}{3}\).
Step 2
Why this answer is correct
\(\alpha=0.60\), अतः \(i=1-\frac{1.2}{3}=0.60\)। / With \(\alpha=0.60\), \(i=1-\frac{1.2}{3}=0.60\).
Step 3
Exam Tip
प्रेक्षित मोलर द्रव्यमान \(=\frac{180}{0.60}=300,g,mol^{-1}\)। / Observed molar mass \(=\frac{180}{0.60}=300,g,mol^{-1}\).
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किसी (0.02,M) विलेय का (300,K) पर परासरण दाब (0.984,atm) है। \(R=0.082,L,atm,mol^{-1},K^{-1}\) हो तो (i) और व्यवहार क्या होगा?
A (0.02,M) solute has osmotic pressure (0.984,atm) at (300,K). If \(R=0.082,L,atm,mol^{-1},K^{-1}\), what are (i) and the behaviour?
#osmotic pressure
#i factor
#dissociation
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A (i=1), सामान्य / (i=1), normal
B (i=2), वियोजन / (i=2), dissociation
C (i=0.5), संघटन / (i=0.5), association
D (i=3), पूर्ण त्रिक वियोजन / (i=3), complete three-particle dissociation
Explanation opens after your attempt
Correct Answer
B. (i=2), वियोजन / (i=2), dissociation
Step 1
Concept
\(i=\frac{\pi}{CRT}=\frac{0.984}{0.02\times0.082\times300}\)। / \(i=\frac{\pi}{CRT}=\frac{0.984}{0.02\times0.082\times300}\).
Step 2
Why this answer is correct
हर (0.492) है, इसलिए (i=2)। / Denominator is (0.492), so (i=2).
Step 3
Exam Tip
(i>1) होने पर वियोजन का संकेत मिलता है। / (i>1) indicates dissociation.
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यदि (0.3,g) विलेय (150,mL) विलयन में (300,K) पर (0.123,atm) परासरण दाब देता है और (i=1), तो मोलर द्रव्यमान क्या होगा?
If (0.3,g) solute in (150,mL) solution gives osmotic pressure (0.123,atm) at (300,K) and (i=1), what is the molar mass?
#osmotic pressure
#large molar mass
#numerical
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A \(300,g,mol^{-1}\)
B \(400,g,mol^{-1}\)
C \(500,g,mol^{-1}\)
D \(600,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
B. \(400,g,mol^{-1}\)
Step 1
Concept
\(C=\frac{0.123}{0.082\times300}=0.005,M\)। / \(C=\frac{0.123}{0.082\times300}=0.005,M\).
Step 2
Why this answer is correct
(150,mL=0.15,L), इसलिए मोल \(0.005\times0.15=0.00075\) हैं। / (150,mL=0.15,L), so moles \(=0.005\times0.15=0.00075\).
Step 3
Exam Tip
मोलर द्रव्यमान \(=\frac{0.3}{0.00075}=400,g,mol^{-1}\)। / Molar mass \(=\frac{0.3}{0.00075}=400,g,mol^{-1}\).
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किसी (AB) विलेय का (i=1.25) है और प्रेक्षित मोलर द्रव्यमान \(96,g,mol^{-1}\) है। वास्तविक मोलर द्रव्यमान और वियोजन की मात्रा क्या होगी?
An (AB) solute has (i=1.25) and observed molar mass \(96,g,mol^{-1}\). What are the true molar mass and degree of dissociation?
#AB electrolyte
#true molar mass
#degree of dissociation
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A \(120,g,mol^{-1}\), (25%)
B \(120,g,mol^{-1}\), (50%)
C \(96,g,mol^{-1}\), (25%)
D \(144,g,mol^{-1}\), (25%)
Explanation opens after your attempt
Correct Answer
A. \(120,g,mol^{-1}\), (25%)
Step 1
Concept
वास्तविक मोलर द्रव्यमान \(=1.25\times96=120,g,mol^{-1}\)। / True molar mass \(=1.25\times96=120,g,mol^{-1}\).
Step 2
Why this answer is correct
(AB) के लिए \(i=1+\alpha\)। / For (AB), \(i=1+\alpha\).
Step 3
Exam Tip
\(\alpha=1.25-1=0.25\), यानी (25%) वियोजन। / \(\alpha=1.25-1=0.25\), meaning (25%) dissociation.
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यदि किसी विलेय का (i=0.4) और वास्तविक मोलर द्रव्यमान \(80,g,mol^{-1}\) है, तो अणुसंख्य विधि से प्रेक्षित मोलर द्रव्यमान क्या होगा?
If a solute has (i=0.4) and true molar mass \(80,g,mol^{-1}\), what will be the observed molar mass from a colligative method?
#association
#observed molar mass
#van't Hoff factor
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A \(32,g,mol^{-1}\)
B \(80,g,mol^{-1}\)
C \(160,g,mol^{-1}\)
D \(200,g,mol^{-1}\)
Explanation opens after your attempt
Correct Answer
D. \(200,g,mol^{-1}\)
Step 1
Concept
\(प्रेक्षित मोलर द्रव्यमान (=\frac{\)वास्तविक मोलर द्रव्यमान}{i})। \(/ Observed molar mass (=\frac{\)true molar mass}{i}).
Step 2
Why this answer is correct
\(\frac{80}{0.4}=200,g,mol^{-1}\)। / \(\frac{80}{0.4}=200,g,mol^{-1}\).
Step 3
Exam Tip
(i<1) होने पर प्रेक्षित मोलर द्रव्यमान वास्तविक से अधिक होता है। / When (i<1), observed molar mass is higher than true mass.
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किसी (0.5,g) विलेय को (250,g) जल में घोलने पर \(\Delta T_f=0.0186,K\) है। यदि (i=1), तो मोलर द्रव्यमान क्या होगा?
When (0.5,g) solute is dissolved in (250,g) water, \(\Delta T_f=0.0186,K\). If (i=1), what is the molar mass?
#small freezing point depression
#molar mass
#cryoscopy
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A \(100,g,mol^{-1}\)
B \(150,g,mol^{-1}\)
C \(200,g,mol^{-1}\)
D \(250,g,mol^{-1}\)
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Correct Answer
C. \(200,g,mol^{-1}\)
Step 1
Concept
\(m=\frac{0.0186}{1.86}=0.01\)। / \(m=\frac{0.0186}{1.86}=0.01\).
Step 2
Why this answer is correct
(250,g=0.25,kg), इसलिए मोल \(0.01\times0.25=0.0025\) हैं। / (250,g=0.25,kg), so moles \(=0.01\times0.25=0.0025\).
Step 3
Exam Tip
मोलर द्रव्यमान \(=\frac{0.5}{0.0025}=200,g,mol^{-1}\)। / Molar mass \(=\frac{0.5}{0.0025}=200,g,mol^{-1}\).
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यदि किसी \(AB_2\) विलेय का (i=2.6) है और वास्तविक मोलर द्रव्यमान \(156,g,mol^{-1}\) है, तो प्रेक्षित मोलर द्रव्यमान और वियोजन की मात्रा क्या होगी?
If an \(AB_2\) solute has (i=2.6) and true molar mass \(156,g,mol^{-1}\), what are the observed molar mass and degree of dissociation?
#AB2 electrolyte
#observed molar mass
#degree of dissociation
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A \(60,g,mol^{-1}\), (80%)
B \(60,g,mol^{-1}\), (60%)
C \(78,g,mol^{-1}\), (80%)
D \(90,g,mol^{-1}\), (40%)
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Correct Answer
A. \(60,g,mol^{-1}\), (80%)
Step 1
Concept
प्रेक्षित मोलर द्रव्यमान \(=\frac{156}{2.6}=60,g,mol^{-1}\)। / Observed molar mass \(=\frac{156}{2.6}=60,g,mol^{-1}\).
Step 2
Why this answer is correct
\(AB_2\) के लिए \(i=1+2\alpha\)। / For \(AB_2\), \(i=1+2\alpha\).
Step 3
Exam Tip
\(2.6=1+2\alpha\), इसलिए \(\alpha=0.8\), यानी (80%)। / \(2.6=1+2\alpha\), so \(\alpha=0.8\), or (80%).
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किसी विलेय के (3.6,g) से (1,L) विलयन बना। (300,K) पर \(\pi=0.369,atm\) है। यदि विलेय (25%) द्विमर बनाता है, तो वास्तविक मोलर द्रव्यमान लगभग क्या होगा?
A (1,L) solution is prepared from (3.6,g) solute. At (300,K), \(\pi=0.369,atm\). If the solute forms dimers to the extent of (25%), what is the approximate true molar mass?
#osmotic pressure
#dimerization
#true molar mass
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A \(200,g,mol^{-1}\)
B \(210,g,mol^{-1}\)
C \(220,g,mol^{-1}\)
D \(240,g,mol^{-1}\)
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Correct Answer
D. \(240,g,mol^{-1}\)
Step 1
Concept
(25%) द्विमर के लिए \(i=1-\frac{0.25}{2}=0.875\)। / For (25%) dimerization, \(i=1-\frac{0.25}{2}=0.875\).
Step 2
Why this answer is correct
\(C=\frac{0.369}{0.875\times0.082\times300}\approx0.0171,M\)। / \(C=\frac{0.369}{0.875\times0.082\times300}\approx0.0171,M\).
Step 3
Exam Tip
(1,L) में मोल (0.0171), इसलिए मोलर द्रव्यमान \(\frac{3.6}{0.0171}\approx210,g,mol^{-1}\)। / In (1,L), moles are (0.0171), so molar mass \(=\frac{3.6}{0.0171}\approx210,g,mol^{-1}\).
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मोलर द्रव्यमान निर्धारण में यदि प्रेक्षित मानों से (i<1) मिलता है, तो सबसे उचित परीक्षा निष्कर्ष क्या होगा?
In molar mass determination, if observed values give (i<1), what is the most suitable exam conclusion?
#conceptual
#molar mass correction
#association
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A विलेय के स्वतंत्र कण बढ़ गए हैं / Independent solute particles have increased
B विलेय के स्वतंत्र कण घटे हैं और संघटन संभव है / Independent solute particles have decreased and association is possible
C विलेय पूर्ण वियोजित है / The solute is completely dissociated
D मोलर द्रव्यमान का कोई संबंध कण संख्या से नहीं है / Molar mass has no relation with particle number
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Correct Answer
B. विलेय के स्वतंत्र कण घटे हैं और संघटन संभव है / Independent solute particles have decreased and association is possible
Step 1
Concept
(i<1) का अर्थ है अणुसंख्य प्रभाव सामान्य से कम है।
Step 2
Why this answer is correct
कम प्रभाव बताता है कि स्वतंत्र विलेय कणों की संख्या घट गई है। / A smaller effect indicates fewer independent solute particles.
Step 3
Exam Tip
ऐसा सामान्यतः संघटन, जैसे द्विमर या त्रिमर बनने, के कारण होता है। / This usually happens due to association, such as dimer or trimer formation.
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