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
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
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
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
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
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
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
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
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
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
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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