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CHXI06:STATES OF MATTER

314133 U-235 and U-238 (as hexa fluorides) can be separated by the method of

1 Chromatography

2 Zone refining

3 Gaseous diffusion

4 Solvent extraction

Explanation:

Due to difference in atomic weight of $U$ isotopes.

CHXI06:STATES OF MATTER

314134 The reaction between gaseous ${NH}_{3}$ and $HBr$ produces a white solid ${NH}_{4} Br$ . Suppose that ${NH}_{3}$ and $HBr$ are introduced simultaneously into the opposite ends of an open tube of 1 metre long. Where would you expect the white solid to form?

1 At a distance of

34.45 cm

from

{NH}_{3}

end

2 At a distance of

68.5 cm

from

{NH}_{3}

end

3 At a distance of

44.45 cm

from

HBr

end

4 At a distance of

45.45 cm

from

HBr

end

Explanation:

Assume, distance from ${NH}_{3}$ end $= x cm$ distance from $HBr$ end $= (1 - x) cm$
$∴ \frac{Distance from {NH}_{3} end}{Distance from HBr end} = \sqrt{\frac{M_{H B r}}{M_{N H_{3}}}}$
$\frac{x}{1 - x} = \sqrt{\frac{81}{17}}$
$x = 68 .5 c m$
$∴$ Distance of $68.5 cm$ from ${NH}_{3}$ end, ${NH}_{4} Br$ formed.

CHXI06:STATES OF MATTER

314135 One mole of nitrogen gas at 0.8 atm takes 38 seconds to diffuse through a pinhole, whereas one mole of an unknown compound of xenon with fluorine at 106 atm takes 57 seconds to diffuse through the same hole. Calculate the molecular mass of the compound.

1 252

2 525

3 262

4 380

Explanation:

$\frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}} \times \frac{P_{1}}{P_{2}} o r \frac{n_{1}}{t_{1}} \times \frac{t_{2}}{n_{2}}$
$= \sqrt{\frac{M_{2}}{M_{1}}} \times \frac{P_{1}}{P_{2}}$
$o r \frac{1}{38} \times \frac{57}{1} = \sqrt{(\frac{M_{g a s}}{28})} \times \frac{0 .8}{1 .6}$
$∴ M_{gas} = {[\frac{57}{38} \times \frac{1.6}{0.8}]}^{2} \times 28 \Rightarrow M_{gas} = 252$

CHXI06:STATES OF MATTER

314136 Pick out the pair of gases with the same rate of diffusion?

1

CO, NO

2

N_{2} O, CO

3

N_{2} O, {CO}_{2}

4

{CO}_{2}, {NO}_{2}

CHXI06:STATES OF MATTER

314137 The ratio of the rate of diffusion of helium and methane under identical conditions of pressure and temperature will be

1

4 : 1

2

2 : 1

3

1 : 2

4

1 : 4

Explanation:

$\frac{r_{He}}{r_{{CH}_{4}}} = \sqrt{\frac{M_{{CH}_{4}}}{M_{He}}} = \sqrt{\frac{16}{4}} = \frac{2}{1} = 2$

CHXI06:STATES OF MATTER

314133 U-235 and U-238 (as hexa fluorides) can be separated by the method of

1 Chromatography

2 Zone refining

3 Gaseous diffusion

4 Solvent extraction

Explanation:

Due to difference in atomic weight of $U$ isotopes.

CHXI06:STATES OF MATTER

314134 The reaction between gaseous ${NH}_{3}$ and $HBr$ produces a white solid ${NH}_{4} Br$ . Suppose that ${NH}_{3}$ and $HBr$ are introduced simultaneously into the opposite ends of an open tube of 1 metre long. Where would you expect the white solid to form?

1 At a distance of

34.45 cm

from

{NH}_{3}

end

2 At a distance of

68.5 cm

from

{NH}_{3}

end

3 At a distance of

44.45 cm

from

HBr

end

4 At a distance of

45.45 cm

from

HBr

end

Explanation:

Assume, distance from ${NH}_{3}$ end $= x cm$ distance from $HBr$ end $= (1 - x) cm$
$∴ \frac{Distance from {NH}_{3} end}{Distance from HBr end} = \sqrt{\frac{M_{H B r}}{M_{N H_{3}}}}$
$\frac{x}{1 - x} = \sqrt{\frac{81}{17}}$
$x = 68 .5 c m$
$∴$ Distance of $68.5 cm$ from ${NH}_{3}$ end, ${NH}_{4} Br$ formed.

CHXI06:STATES OF MATTER

314135 One mole of nitrogen gas at 0.8 atm takes 38 seconds to diffuse through a pinhole, whereas one mole of an unknown compound of xenon with fluorine at 106 atm takes 57 seconds to diffuse through the same hole. Calculate the molecular mass of the compound.

1 252

2 525

3 262

4 380

Explanation:

$\frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}} \times \frac{P_{1}}{P_{2}} o r \frac{n_{1}}{t_{1}} \times \frac{t_{2}}{n_{2}}$
$= \sqrt{\frac{M_{2}}{M_{1}}} \times \frac{P_{1}}{P_{2}}$
$o r \frac{1}{38} \times \frac{57}{1} = \sqrt{(\frac{M_{g a s}}{28})} \times \frac{0 .8}{1 .6}$
$∴ M_{gas} = {[\frac{57}{38} \times \frac{1.6}{0.8}]}^{2} \times 28 \Rightarrow M_{gas} = 252$

CHXI06:STATES OF MATTER

314136 Pick out the pair of gases with the same rate of diffusion?

1

CO, NO

2

N_{2} O, CO

3

N_{2} O, {CO}_{2}

4

{CO}_{2}, {NO}_{2}

CHXI06:STATES OF MATTER

314137 The ratio of the rate of diffusion of helium and methane under identical conditions of pressure and temperature will be

1

4 : 1

2

2 : 1

3

1 : 2

4

1 : 4

Explanation:

$\frac{r_{He}}{r_{{CH}_{4}}} = \sqrt{\frac{M_{{CH}_{4}}}{M_{He}}} = \sqrt{\frac{16}{4}} = \frac{2}{1} = 2$

CHXI06:STATES OF MATTER

314133 U-235 and U-238 (as hexa fluorides) can be separated by the method of

1 Chromatography

2 Zone refining

3 Gaseous diffusion

4 Solvent extraction

Explanation:

Due to difference in atomic weight of $U$ isotopes.

CHXI06:STATES OF MATTER

314134 The reaction between gaseous ${NH}_{3}$ and $HBr$ produces a white solid ${NH}_{4} Br$ . Suppose that ${NH}_{3}$ and $HBr$ are introduced simultaneously into the opposite ends of an open tube of 1 metre long. Where would you expect the white solid to form?

1 At a distance of

34.45 cm

from

{NH}_{3}

end

2 At a distance of

68.5 cm

from

{NH}_{3}

end

3 At a distance of

44.45 cm

from

HBr

end

4 At a distance of

45.45 cm

from

HBr

end

Explanation:

Assume, distance from ${NH}_{3}$ end $= x cm$ distance from $HBr$ end $= (1 - x) cm$
$∴ \frac{Distance from {NH}_{3} end}{Distance from HBr end} = \sqrt{\frac{M_{H B r}}{M_{N H_{3}}}}$
$\frac{x}{1 - x} = \sqrt{\frac{81}{17}}$
$x = 68 .5 c m$
$∴$ Distance of $68.5 cm$ from ${NH}_{3}$ end, ${NH}_{4} Br$ formed.

CHXI06:STATES OF MATTER

314135 One mole of nitrogen gas at 0.8 atm takes 38 seconds to diffuse through a pinhole, whereas one mole of an unknown compound of xenon with fluorine at 106 atm takes 57 seconds to diffuse through the same hole. Calculate the molecular mass of the compound.

1 252

2 525

3 262

4 380

Explanation:

$\frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}} \times \frac{P_{1}}{P_{2}} o r \frac{n_{1}}{t_{1}} \times \frac{t_{2}}{n_{2}}$
$= \sqrt{\frac{M_{2}}{M_{1}}} \times \frac{P_{1}}{P_{2}}$
$o r \frac{1}{38} \times \frac{57}{1} = \sqrt{(\frac{M_{g a s}}{28})} \times \frac{0 .8}{1 .6}$
$∴ M_{gas} = {[\frac{57}{38} \times \frac{1.6}{0.8}]}^{2} \times 28 \Rightarrow M_{gas} = 252$

CHXI06:STATES OF MATTER

314136 Pick out the pair of gases with the same rate of diffusion?

1

CO, NO

2

N_{2} O, CO

3

N_{2} O, {CO}_{2}

4

{CO}_{2}, {NO}_{2}

CHXI06:STATES OF MATTER

314137 The ratio of the rate of diffusion of helium and methane under identical conditions of pressure and temperature will be

1

4 : 1

2

2 : 1

3

1 : 2

4

1 : 4

Explanation:

$\frac{r_{He}}{r_{{CH}_{4}}} = \sqrt{\frac{M_{{CH}_{4}}}{M_{He}}} = \sqrt{\frac{16}{4}} = \frac{2}{1} = 2$

CHXI06:STATES OF MATTER

314133 U-235 and U-238 (as hexa fluorides) can be separated by the method of

1 Chromatography

2 Zone refining

3 Gaseous diffusion

4 Solvent extraction

Explanation:

Due to difference in atomic weight of $U$ isotopes.

CHXI06:STATES OF MATTER

314134 The reaction between gaseous ${NH}_{3}$ and $HBr$ produces a white solid ${NH}_{4} Br$ . Suppose that ${NH}_{3}$ and $HBr$ are introduced simultaneously into the opposite ends of an open tube of 1 metre long. Where would you expect the white solid to form?

1 At a distance of

34.45 cm

from

{NH}_{3}

end

2 At a distance of

68.5 cm

from

{NH}_{3}

end

3 At a distance of

44.45 cm

from

HBr

end

4 At a distance of

45.45 cm

from

HBr

end

Explanation:

Assume, distance from ${NH}_{3}$ end $= x cm$ distance from $HBr$ end $= (1 - x) cm$
$∴ \frac{Distance from {NH}_{3} end}{Distance from HBr end} = \sqrt{\frac{M_{H B r}}{M_{N H_{3}}}}$
$\frac{x}{1 - x} = \sqrt{\frac{81}{17}}$
$x = 68 .5 c m$
$∴$ Distance of $68.5 cm$ from ${NH}_{3}$ end, ${NH}_{4} Br$ formed.

CHXI06:STATES OF MATTER

314135 One mole of nitrogen gas at 0.8 atm takes 38 seconds to diffuse through a pinhole, whereas one mole of an unknown compound of xenon with fluorine at 106 atm takes 57 seconds to diffuse through the same hole. Calculate the molecular mass of the compound.

1 252

2 525

3 262

4 380

Explanation:

$\frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}} \times \frac{P_{1}}{P_{2}} o r \frac{n_{1}}{t_{1}} \times \frac{t_{2}}{n_{2}}$
$= \sqrt{\frac{M_{2}}{M_{1}}} \times \frac{P_{1}}{P_{2}}$
$o r \frac{1}{38} \times \frac{57}{1} = \sqrt{(\frac{M_{g a s}}{28})} \times \frac{0 .8}{1 .6}$
$∴ M_{gas} = {[\frac{57}{38} \times \frac{1.6}{0.8}]}^{2} \times 28 \Rightarrow M_{gas} = 252$

CHXI06:STATES OF MATTER

314136 Pick out the pair of gases with the same rate of diffusion?

1

CO, NO

2

N_{2} O, CO

3

N_{2} O, {CO}_{2}

4

{CO}_{2}, {NO}_{2}

CHXI06:STATES OF MATTER

314137 The ratio of the rate of diffusion of helium and methane under identical conditions of pressure and temperature will be

1

4 : 1

2

2 : 1

3

1 : 2

4

1 : 4

Explanation:

$\frac{r_{He}}{r_{{CH}_{4}}} = \sqrt{\frac{M_{{CH}_{4}}}{M_{He}}} = \sqrt{\frac{16}{4}} = \frac{2}{1} = 2$

CHXI06:STATES OF MATTER

314133 U-235 and U-238 (as hexa fluorides) can be separated by the method of

1 Chromatography

2 Zone refining

3 Gaseous diffusion

4 Solvent extraction

Explanation:

Due to difference in atomic weight of $U$ isotopes.

CHXI06:STATES OF MATTER

314134 The reaction between gaseous ${NH}_{3}$ and $HBr$ produces a white solid ${NH}_{4} Br$ . Suppose that ${NH}_{3}$ and $HBr$ are introduced simultaneously into the opposite ends of an open tube of 1 metre long. Where would you expect the white solid to form?

1 At a distance of

34.45 cm

from

{NH}_{3}

end

2 At a distance of

68.5 cm

from

{NH}_{3}

end

3 At a distance of

44.45 cm

from

HBr

end

4 At a distance of

45.45 cm

from

HBr

end

Explanation:

Assume, distance from ${NH}_{3}$ end $= x cm$ distance from $HBr$ end $= (1 - x) cm$
$∴ \frac{Distance from {NH}_{3} end}{Distance from HBr end} = \sqrt{\frac{M_{H B r}}{M_{N H_{3}}}}$
$\frac{x}{1 - x} = \sqrt{\frac{81}{17}}$
$x = 68 .5 c m$
$∴$ Distance of $68.5 cm$ from ${NH}_{3}$ end, ${NH}_{4} Br$ formed.

CHXI06:STATES OF MATTER

314135 One mole of nitrogen gas at 0.8 atm takes 38 seconds to diffuse through a pinhole, whereas one mole of an unknown compound of xenon with fluorine at 106 atm takes 57 seconds to diffuse through the same hole. Calculate the molecular mass of the compound.

1 252

2 525

3 262

4 380

Explanation:

$\frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}} \times \frac{P_{1}}{P_{2}} o r \frac{n_{1}}{t_{1}} \times \frac{t_{2}}{n_{2}}$
$= \sqrt{\frac{M_{2}}{M_{1}}} \times \frac{P_{1}}{P_{2}}$
$o r \frac{1}{38} \times \frac{57}{1} = \sqrt{(\frac{M_{g a s}}{28})} \times \frac{0 .8}{1 .6}$
$∴ M_{gas} = {[\frac{57}{38} \times \frac{1.6}{0.8}]}^{2} \times 28 \Rightarrow M_{gas} = 252$

CHXI06:STATES OF MATTER

314136 Pick out the pair of gases with the same rate of diffusion?

1

CO, NO

2

N_{2} O, CO

3

N_{2} O, {CO}_{2}

4

{CO}_{2}, {NO}_{2}

CHXI06:STATES OF MATTER

314137 The ratio of the rate of diffusion of helium and methane under identical conditions of pressure and temperature will be

1

4 : 1

2

2 : 1

3

1 : 2

4

1 : 4

Explanation:

$\frac{r_{He}}{r_{{CH}_{4}}} = \sqrt{\frac{M_{{CH}_{4}}}{M_{He}}} = \sqrt{\frac{16}{4}} = \frac{2}{1} = 2$

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