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Chemistry 6: Molar Mass Determination MCQ Questions for Class 12 General

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6: Molar Mass Determination Practice Questions

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Question 491/587 Expert Chemistry Chapter 01: Solutions 6: Molar Mass Determination Class 12 Level 17

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?

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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Question 492/587 Expert Chemistry Chapter 01: Solutions 6: Molar Mass Determination Class 12 Level 17

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?

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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Question 493/587 Expert Chemistry Chapter 01: Solutions 6: Molar Mass Determination Class 12 Level 17

\(MgCl_2\) 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?

Explanation opens after your attempt

Correct Answer

B. (50%)

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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Question 494/587 Expert Chemistry Chapter 01: Solutions 6: Molar Mass Determination Class 12 Level 17

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?

Explanation opens after your attempt

Correct Answer

C. (40%)

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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Question 495/587 Expert Chemistry Chapter 01: Solutions 6: Molar Mass Determination Class 12 Level 17

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?

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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Question 496/587 Expert Chemistry Chapter 01: Solutions 6: Molar Mass Determination Class 12 Level 17

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?

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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Question 497/587 Expert Chemistry Chapter 01: Solutions 6: Molar Mass Determination Class 12 Level 17

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?

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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Question 498/587 Expert Chemistry Chapter 01: Solutions 6: Molar Mass Determination Class 12 Level 17

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?

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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Question 499/587 Expert Chemistry Chapter 01: Solutions 6: Molar Mass Determination Class 12 Level 17

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?

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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Question 500/587 Expert Chemistry Chapter 01: Solutions 6: Molar Mass Determination Class 12 Level 17

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?

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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Chemistry 6: Molar Mass Determination MCQ Questions for Class 12 General

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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?

Concept

Why this answer is correct

Exam Tip

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?

Concept

Why this answer is correct

Exam Tip

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?

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