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CHXI06:STATES OF MATTER

314146 The odour from a gas A takes 6 seconds to reach a wall from a given point. If the molecular weight of gas A is 46 grams per mole how long will it take for the odour from gas B to reach the same wall from the same point? Assuming that volume of both gases is same.

1

6 \sec

2

7 \sec

3

8 \sec

4

9 \sec

Explanation:

$\frac{r_{A}}{r_{B}} = \sqrt{\frac{M_{B}}{M_{A}}}$ (from Graham's law of diffusion)
$\frac{r_{A}}{r_{B}} = \sqrt{\frac{64}{46}} = 1 .18 \Rightarrow \frac{V_{A} / t_{A}}{V_{B} / t_{B}} = 1 .18$
$∴ t_{B} = 1 .18 \times t_{A}$
Time taken for the odour of B to reach the wall $(t_{B}) = 1.18 \times 6 = 7.08 \sec ≅ 7 \sec$ .

CHXI06:STATES OF MATTER

314147 Sulphur dioxide and oxygen were allowed to diffuse through a porous partition. $20 {dm}^{3}$ of ${SO}_{2}$ diffuses through the porous partition in 60 seconds. The volume of $O_{2}$ in ${dm}^{3}$ which diffuses under the similar condition in 30 seconds will be (atomic mass of sulphur $= 32 u$ ):

1 7.09

2 14.1

3 10.0

4 28.2

Explanation:

According to Graham's Law Diffusion:
$\frac{r_{1}}{r_{2}} = \sqrt{\frac{d_{2}}{d_{1}}}$
Since rate of diffusion
$= \frac{Vol. of gas diffused (V)}{Time taken for diffusion (t)}$
$∴ \frac{r_{1}}{r_{2}} = \frac{V_{1} / t_{1}}{V_{2} / t_{2}}$
or $\frac{r_{1}}{r_{2}} = \frac{V_{1} / t_{1}}{V_{2} / t_{2}} = \sqrt{\frac{d_{2}}{d_{1}}}$
$∴ \frac{20 / 60}{V_{2} / 30} = \sqrt{\frac{16 / 2}{32 / 2}} = \sqrt{\frac{1}{2}}$
$∵$ Mol. wt $= 2 \times V . D$
$∴ V . D = \frac{M o l . w t}{2}$
On calculating,
$V_{2} = 14 .1$

JEE - 2014

CHXI06:STATES OF MATTER

314148 Four gas balloons $A, B, C, D$ of equal volumes containing $H_{2}, N_{2} O, CO, {CO}_{2}$ respectively were picked with needle and immersed in a tank containing ${CO}_{2}$ . Which of them will shrink after some time?

1 All

2 Both A and C

3 Only C

4 Both A and D

CHXI06:STATES OF MATTER

314149 At what temperature will the rate of diffusion of $N_{2}$ be 1.6 times the rate of diffusion of ${SO}_{2}$ at $27^{\circ} C$ ?

1

336^{\circ} C

2

300 K

3

50^{\circ} C

4

63^{\circ} C

Explanation:

Given, $r_{N_{2}} = 1.6 r_{{SO}_{2}}$
$\frac{r_{N_{2}}}{r_{S O_{2}}} = 1 .6$
$∴ \frac{r_{N_{2}}}{r_{S O_{2}}} = \sqrt{\frac{T_{N_{2}}}{T_{S O_{2}}} \times \frac{M_{S O_{2}}}{M_{N_{2}}}} 1 .6$
$= \sqrt{\frac{T_{N_{2}}}{300} \times \frac{64}{28}}$
$T_{N_{2}} = 336 K = 6 3^{^{\circ}} C$

CHXI06:STATES OF MATTER

314150 A gaseous mixture containing $He, {CH}_{4}$ , and ${SO}_{2}$ was allowed to effuse through a fine hole. Find the molar ratio of the gases coming out initially if the mixture contains $He, {CH}_{4}$ and ${SO}_{2}$ in 1:2:3 mole ratio

1

2 : 2 : 3

2

6 : 6 : 1

3

\sqrt{2} : \sqrt{2} : 3

4

4 : 4 : 3

Explanation:

$\frac{n_{H e}^{'}}{n_{C H_{4}}^{'}} = \frac{1}{2} \sqrt{\frac{16}{4}} = \frac{1}{1}$
$\frac{n_{He}^{'}}{n_{CO 2}^{'}} = \frac{1}{3} \sqrt{\frac{64}{4}} = \frac{4}{3}$
$n_{H e}^{'} : n_{C H_{4}}^{'} : n_{S_{2}}^{'} = 4 : 4 : 3$

CHXI06:STATES OF MATTER

314146 The odour from a gas A takes 6 seconds to reach a wall from a given point. If the molecular weight of gas A is 46 grams per mole how long will it take for the odour from gas B to reach the same wall from the same point? Assuming that volume of both gases is same.

1

6 \sec

2

7 \sec

3

8 \sec

4

9 \sec

Explanation:

$\frac{r_{A}}{r_{B}} = \sqrt{\frac{M_{B}}{M_{A}}}$ (from Graham's law of diffusion)
$\frac{r_{A}}{r_{B}} = \sqrt{\frac{64}{46}} = 1 .18 \Rightarrow \frac{V_{A} / t_{A}}{V_{B} / t_{B}} = 1 .18$
$∴ t_{B} = 1 .18 \times t_{A}$
Time taken for the odour of B to reach the wall $(t_{B}) = 1.18 \times 6 = 7.08 \sec ≅ 7 \sec$ .

CHXI06:STATES OF MATTER

314147 Sulphur dioxide and oxygen were allowed to diffuse through a porous partition. $20 {dm}^{3}$ of ${SO}_{2}$ diffuses through the porous partition in 60 seconds. The volume of $O_{2}$ in ${dm}^{3}$ which diffuses under the similar condition in 30 seconds will be (atomic mass of sulphur $= 32 u$ ):

1 7.09

2 14.1

3 10.0

4 28.2

Explanation:

According to Graham's Law Diffusion:
$\frac{r_{1}}{r_{2}} = \sqrt{\frac{d_{2}}{d_{1}}}$
Since rate of diffusion
$= \frac{Vol. of gas diffused (V)}{Time taken for diffusion (t)}$
$∴ \frac{r_{1}}{r_{2}} = \frac{V_{1} / t_{1}}{V_{2} / t_{2}}$
or $\frac{r_{1}}{r_{2}} = \frac{V_{1} / t_{1}}{V_{2} / t_{2}} = \sqrt{\frac{d_{2}}{d_{1}}}$
$∴ \frac{20 / 60}{V_{2} / 30} = \sqrt{\frac{16 / 2}{32 / 2}} = \sqrt{\frac{1}{2}}$
$∵$ Mol. wt $= 2 \times V . D$
$∴ V . D = \frac{M o l . w t}{2}$
On calculating,
$V_{2} = 14 .1$

JEE - 2014

CHXI06:STATES OF MATTER

314148 Four gas balloons $A, B, C, D$ of equal volumes containing $H_{2}, N_{2} O, CO, {CO}_{2}$ respectively were picked with needle and immersed in a tank containing ${CO}_{2}$ . Which of them will shrink after some time?

1 All

2 Both A and C

3 Only C

4 Both A and D

CHXI06:STATES OF MATTER

314149 At what temperature will the rate of diffusion of $N_{2}$ be 1.6 times the rate of diffusion of ${SO}_{2}$ at $27^{\circ} C$ ?

1

336^{\circ} C

2

300 K

3

50^{\circ} C

4

63^{\circ} C

Explanation:

Given, $r_{N_{2}} = 1.6 r_{{SO}_{2}}$
$\frac{r_{N_{2}}}{r_{S O_{2}}} = 1 .6$
$∴ \frac{r_{N_{2}}}{r_{S O_{2}}} = \sqrt{\frac{T_{N_{2}}}{T_{S O_{2}}} \times \frac{M_{S O_{2}}}{M_{N_{2}}}} 1 .6$
$= \sqrt{\frac{T_{N_{2}}}{300} \times \frac{64}{28}}$
$T_{N_{2}} = 336 K = 6 3^{^{\circ}} C$

CHXI06:STATES OF MATTER

314150 A gaseous mixture containing $He, {CH}_{4}$ , and ${SO}_{2}$ was allowed to effuse through a fine hole. Find the molar ratio of the gases coming out initially if the mixture contains $He, {CH}_{4}$ and ${SO}_{2}$ in 1:2:3 mole ratio

1

2 : 2 : 3

2

6 : 6 : 1

3

\sqrt{2} : \sqrt{2} : 3

4

4 : 4 : 3

Explanation:

$\frac{n_{H e}^{'}}{n_{C H_{4}}^{'}} = \frac{1}{2} \sqrt{\frac{16}{4}} = \frac{1}{1}$
$\frac{n_{He}^{'}}{n_{CO 2}^{'}} = \frac{1}{3} \sqrt{\frac{64}{4}} = \frac{4}{3}$
$n_{H e}^{'} : n_{C H_{4}}^{'} : n_{S_{2}}^{'} = 4 : 4 : 3$

CHXI06:STATES OF MATTER

314146 The odour from a gas A takes 6 seconds to reach a wall from a given point. If the molecular weight of gas A is 46 grams per mole how long will it take for the odour from gas B to reach the same wall from the same point? Assuming that volume of both gases is same.

1

6 \sec

2

7 \sec

3

8 \sec

4

9 \sec

Explanation:

$\frac{r_{A}}{r_{B}} = \sqrt{\frac{M_{B}}{M_{A}}}$ (from Graham's law of diffusion)
$\frac{r_{A}}{r_{B}} = \sqrt{\frac{64}{46}} = 1 .18 \Rightarrow \frac{V_{A} / t_{A}}{V_{B} / t_{B}} = 1 .18$
$∴ t_{B} = 1 .18 \times t_{A}$
Time taken for the odour of B to reach the wall $(t_{B}) = 1.18 \times 6 = 7.08 \sec ≅ 7 \sec$ .

CHXI06:STATES OF MATTER

314147 Sulphur dioxide and oxygen were allowed to diffuse through a porous partition. $20 {dm}^{3}$ of ${SO}_{2}$ diffuses through the porous partition in 60 seconds. The volume of $O_{2}$ in ${dm}^{3}$ which diffuses under the similar condition in 30 seconds will be (atomic mass of sulphur $= 32 u$ ):

1 7.09

2 14.1

3 10.0

4 28.2

Explanation:

According to Graham's Law Diffusion:
$\frac{r_{1}}{r_{2}} = \sqrt{\frac{d_{2}}{d_{1}}}$
Since rate of diffusion
$= \frac{Vol. of gas diffused (V)}{Time taken for diffusion (t)}$
$∴ \frac{r_{1}}{r_{2}} = \frac{V_{1} / t_{1}}{V_{2} / t_{2}}$
or $\frac{r_{1}}{r_{2}} = \frac{V_{1} / t_{1}}{V_{2} / t_{2}} = \sqrt{\frac{d_{2}}{d_{1}}}$
$∴ \frac{20 / 60}{V_{2} / 30} = \sqrt{\frac{16 / 2}{32 / 2}} = \sqrt{\frac{1}{2}}$
$∵$ Mol. wt $= 2 \times V . D$
$∴ V . D = \frac{M o l . w t}{2}$
On calculating,
$V_{2} = 14 .1$

JEE - 2014

CHXI06:STATES OF MATTER

314148 Four gas balloons $A, B, C, D$ of equal volumes containing $H_{2}, N_{2} O, CO, {CO}_{2}$ respectively were picked with needle and immersed in a tank containing ${CO}_{2}$ . Which of them will shrink after some time?

1 All

2 Both A and C

3 Only C

4 Both A and D

CHXI06:STATES OF MATTER

314149 At what temperature will the rate of diffusion of $N_{2}$ be 1.6 times the rate of diffusion of ${SO}_{2}$ at $27^{\circ} C$ ?

1

336^{\circ} C

2

300 K

3

50^{\circ} C

4

63^{\circ} C

Explanation:

Given, $r_{N_{2}} = 1.6 r_{{SO}_{2}}$
$\frac{r_{N_{2}}}{r_{S O_{2}}} = 1 .6$
$∴ \frac{r_{N_{2}}}{r_{S O_{2}}} = \sqrt{\frac{T_{N_{2}}}{T_{S O_{2}}} \times \frac{M_{S O_{2}}}{M_{N_{2}}}} 1 .6$
$= \sqrt{\frac{T_{N_{2}}}{300} \times \frac{64}{28}}$
$T_{N_{2}} = 336 K = 6 3^{^{\circ}} C$

CHXI06:STATES OF MATTER

314150 A gaseous mixture containing $He, {CH}_{4}$ , and ${SO}_{2}$ was allowed to effuse through a fine hole. Find the molar ratio of the gases coming out initially if the mixture contains $He, {CH}_{4}$ and ${SO}_{2}$ in 1:2:3 mole ratio

1

2 : 2 : 3

2

6 : 6 : 1

3

\sqrt{2} : \sqrt{2} : 3

4

4 : 4 : 3

Explanation:

$\frac{n_{H e}^{'}}{n_{C H_{4}}^{'}} = \frac{1}{2} \sqrt{\frac{16}{4}} = \frac{1}{1}$
$\frac{n_{He}^{'}}{n_{CO 2}^{'}} = \frac{1}{3} \sqrt{\frac{64}{4}} = \frac{4}{3}$
$n_{H e}^{'} : n_{C H_{4}}^{'} : n_{S_{2}}^{'} = 4 : 4 : 3$

CHXI06:STATES OF MATTER

314146 The odour from a gas A takes 6 seconds to reach a wall from a given point. If the molecular weight of gas A is 46 grams per mole how long will it take for the odour from gas B to reach the same wall from the same point? Assuming that volume of both gases is same.

1

6 \sec

2

7 \sec

3

8 \sec

4

9 \sec

Explanation:

$\frac{r_{A}}{r_{B}} = \sqrt{\frac{M_{B}}{M_{A}}}$ (from Graham's law of diffusion)
$\frac{r_{A}}{r_{B}} = \sqrt{\frac{64}{46}} = 1 .18 \Rightarrow \frac{V_{A} / t_{A}}{V_{B} / t_{B}} = 1 .18$
$∴ t_{B} = 1 .18 \times t_{A}$
Time taken for the odour of B to reach the wall $(t_{B}) = 1.18 \times 6 = 7.08 \sec ≅ 7 \sec$ .

CHXI06:STATES OF MATTER

314147 Sulphur dioxide and oxygen were allowed to diffuse through a porous partition. $20 {dm}^{3}$ of ${SO}_{2}$ diffuses through the porous partition in 60 seconds. The volume of $O_{2}$ in ${dm}^{3}$ which diffuses under the similar condition in 30 seconds will be (atomic mass of sulphur $= 32 u$ ):

1 7.09

2 14.1

3 10.0

4 28.2

Explanation:

According to Graham's Law Diffusion:
$\frac{r_{1}}{r_{2}} = \sqrt{\frac{d_{2}}{d_{1}}}$
Since rate of diffusion
$= \frac{Vol. of gas diffused (V)}{Time taken for diffusion (t)}$
$∴ \frac{r_{1}}{r_{2}} = \frac{V_{1} / t_{1}}{V_{2} / t_{2}}$
or $\frac{r_{1}}{r_{2}} = \frac{V_{1} / t_{1}}{V_{2} / t_{2}} = \sqrt{\frac{d_{2}}{d_{1}}}$
$∴ \frac{20 / 60}{V_{2} / 30} = \sqrt{\frac{16 / 2}{32 / 2}} = \sqrt{\frac{1}{2}}$
$∵$ Mol. wt $= 2 \times V . D$
$∴ V . D = \frac{M o l . w t}{2}$
On calculating,
$V_{2} = 14 .1$

JEE - 2014

CHXI06:STATES OF MATTER

314148 Four gas balloons $A, B, C, D$ of equal volumes containing $H_{2}, N_{2} O, CO, {CO}_{2}$ respectively were picked with needle and immersed in a tank containing ${CO}_{2}$ . Which of them will shrink after some time?

1 All

2 Both A and C

3 Only C

4 Both A and D

CHXI06:STATES OF MATTER

314149 At what temperature will the rate of diffusion of $N_{2}$ be 1.6 times the rate of diffusion of ${SO}_{2}$ at $27^{\circ} C$ ?

1

336^{\circ} C

2

300 K

3

50^{\circ} C

4

63^{\circ} C

Explanation:

Given, $r_{N_{2}} = 1.6 r_{{SO}_{2}}$
$\frac{r_{N_{2}}}{r_{S O_{2}}} = 1 .6$
$∴ \frac{r_{N_{2}}}{r_{S O_{2}}} = \sqrt{\frac{T_{N_{2}}}{T_{S O_{2}}} \times \frac{M_{S O_{2}}}{M_{N_{2}}}} 1 .6$
$= \sqrt{\frac{T_{N_{2}}}{300} \times \frac{64}{28}}$
$T_{N_{2}} = 336 K = 6 3^{^{\circ}} C$

CHXI06:STATES OF MATTER

314150 A gaseous mixture containing $He, {CH}_{4}$ , and ${SO}_{2}$ was allowed to effuse through a fine hole. Find the molar ratio of the gases coming out initially if the mixture contains $He, {CH}_{4}$ and ${SO}_{2}$ in 1:2:3 mole ratio

1

2 : 2 : 3

2

6 : 6 : 1

3

\sqrt{2} : \sqrt{2} : 3

4

4 : 4 : 3

Explanation:

$\frac{n_{H e}^{'}}{n_{C H_{4}}^{'}} = \frac{1}{2} \sqrt{\frac{16}{4}} = \frac{1}{1}$
$\frac{n_{He}^{'}}{n_{CO 2}^{'}} = \frac{1}{3} \sqrt{\frac{64}{4}} = \frac{4}{3}$
$n_{H e}^{'} : n_{C H_{4}}^{'} : n_{S_{2}}^{'} = 4 : 4 : 3$

CHXI06:STATES OF MATTER

314146 The odour from a gas A takes 6 seconds to reach a wall from a given point. If the molecular weight of gas A is 46 grams per mole how long will it take for the odour from gas B to reach the same wall from the same point? Assuming that volume of both gases is same.

1

6 \sec

2

7 \sec

3

8 \sec

4

9 \sec

Explanation:

$\frac{r_{A}}{r_{B}} = \sqrt{\frac{M_{B}}{M_{A}}}$ (from Graham's law of diffusion)
$\frac{r_{A}}{r_{B}} = \sqrt{\frac{64}{46}} = 1 .18 \Rightarrow \frac{V_{A} / t_{A}}{V_{B} / t_{B}} = 1 .18$
$∴ t_{B} = 1 .18 \times t_{A}$
Time taken for the odour of B to reach the wall $(t_{B}) = 1.18 \times 6 = 7.08 \sec ≅ 7 \sec$ .

CHXI06:STATES OF MATTER

314147 Sulphur dioxide and oxygen were allowed to diffuse through a porous partition. $20 {dm}^{3}$ of ${SO}_{2}$ diffuses through the porous partition in 60 seconds. The volume of $O_{2}$ in ${dm}^{3}$ which diffuses under the similar condition in 30 seconds will be (atomic mass of sulphur $= 32 u$ ):

1 7.09

2 14.1

3 10.0

4 28.2

Explanation:

According to Graham's Law Diffusion:
$\frac{r_{1}}{r_{2}} = \sqrt{\frac{d_{2}}{d_{1}}}$
Since rate of diffusion
$= \frac{Vol. of gas diffused (V)}{Time taken for diffusion (t)}$
$∴ \frac{r_{1}}{r_{2}} = \frac{V_{1} / t_{1}}{V_{2} / t_{2}}$
or $\frac{r_{1}}{r_{2}} = \frac{V_{1} / t_{1}}{V_{2} / t_{2}} = \sqrt{\frac{d_{2}}{d_{1}}}$
$∴ \frac{20 / 60}{V_{2} / 30} = \sqrt{\frac{16 / 2}{32 / 2}} = \sqrt{\frac{1}{2}}$
$∵$ Mol. wt $= 2 \times V . D$
$∴ V . D = \frac{M o l . w t}{2}$
On calculating,
$V_{2} = 14 .1$

JEE - 2014

CHXI06:STATES OF MATTER

314148 Four gas balloons $A, B, C, D$ of equal volumes containing $H_{2}, N_{2} O, CO, {CO}_{2}$ respectively were picked with needle and immersed in a tank containing ${CO}_{2}$ . Which of them will shrink after some time?

1 All

2 Both A and C

3 Only C

4 Both A and D

CHXI06:STATES OF MATTER

314149 At what temperature will the rate of diffusion of $N_{2}$ be 1.6 times the rate of diffusion of ${SO}_{2}$ at $27^{\circ} C$ ?

1

336^{\circ} C

2

300 K

3

50^{\circ} C

4

63^{\circ} C

Explanation:

Given, $r_{N_{2}} = 1.6 r_{{SO}_{2}}$
$\frac{r_{N_{2}}}{r_{S O_{2}}} = 1 .6$
$∴ \frac{r_{N_{2}}}{r_{S O_{2}}} = \sqrt{\frac{T_{N_{2}}}{T_{S O_{2}}} \times \frac{M_{S O_{2}}}{M_{N_{2}}}} 1 .6$
$= \sqrt{\frac{T_{N_{2}}}{300} \times \frac{64}{28}}$
$T_{N_{2}} = 336 K = 6 3^{^{\circ}} C$

CHXI06:STATES OF MATTER

314150 A gaseous mixture containing $He, {CH}_{4}$ , and ${SO}_{2}$ was allowed to effuse through a fine hole. Find the molar ratio of the gases coming out initially if the mixture contains $He, {CH}_{4}$ and ${SO}_{2}$ in 1:2:3 mole ratio

1

2 : 2 : 3

2

6 : 6 : 1

3

\sqrt{2} : \sqrt{2} : 3

4

4 : 4 : 3

Explanation:

$\frac{n_{H e}^{'}}{n_{C H_{4}}^{'}} = \frac{1}{2} \sqrt{\frac{16}{4}} = \frac{1}{1}$
$\frac{n_{He}^{'}}{n_{CO 2}^{'}} = \frac{1}{3} \sqrt{\frac{64}{4}} = \frac{4}{3}$
$n_{H e}^{'} : n_{C H_{4}}^{'} : n_{S_{2}}^{'} = 4 : 4 : 3$