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CHXI06:STATES OF MATTER

314104 Which gas cannot be collected over water?

1

{NH}_{3}

2

CO

3

N_{2}

4

H_{2}

CHXI06:STATES OF MATTER

314105 $2 .8 g o f N_{2}, 2 .8 g C O$ , and $4 .4 o f C O_{2}$ are found to exert a pressure of 700 torr. Find partial pressure of $N_{2}$ gas in the mixture.

1 280.8 torr

2 233.3 torr

3 300 torr

4 350 torr

Explanation:

$P_{N_{2}} =$ Mole fraction of $N_{2} \times P_{total}$
$= \frac{2 .8 / 28}{\frac{2 .8}{28} + \frac{2 .8}{28} + \frac{4 .4}{44}} \times 700$
$= \frac{0 .1}{0 .3} \times 700$
$= 233 .3 t o r r .$

CHXI06:STATES OF MATTER

314106 Equal weights of ${CH}_{4}$ and $H_{2}$ are mixed in a container at $25^{\circ} C$ . Fraction of total pressure exerted by methane is

1

\frac{1}{2}

2

\frac{1}{3}

3

\frac{1}{9}

4

\frac{8}{9}

Explanation:

Suppose w kg of each is taken
$n_{C H_{4}} = w / 16 a n d n_{H_{2}} = w / 2$
$p_{C H_{4}} = p_{t o t a l} \times$ mole fraction
$= p_{t o t a l} \times \frac{w / 16}{w / 2 + w / 16}$
$p_{C H_{4}} = p_{t o t a l} \times 1 / 9$
$∴$ Fraction of total pressure exerted by methane is $1 / 9$ .

CHXI06:STATES OF MATTER

314107 In a ten litre vessel, the total pressure of a gaseous mixture containing $H_{2}, N_{2}$ and ${CO}_{2}$ is 9.8 atm. The partial pressures of $H_{2}$ and $N_{2}$ are 3.7 and $4.2 atm$ respectively. Then the partial pressure of ${CO}_{2}$ is

1

1.9 atm

2

0.19 atm

3

2.4 atm

4

0.019 atm

Explanation:

$P_{t o t a l} = P (N_{2}) + P (H_{2}) + P (C O_{2})$
$9 .8 = 3 .7 + 4 .2 + P (C O_{2})$
$P (C O_{2}) = 1 .9 a t m$

CHXI06:STATES OF MATTER

314108 A gaseous mixture was prepared by taking equal moles of $CO$ and $N_{2}$ . If the total pressure of the mixture was found to be 1 atmosphere, the partial pressure of the nitrogen $(N_{2})$ in the mixture is

1

1 atm

2

0.9 atm

3

0.8 atm

4

0.5 atm

Explanation:

Given, $n_{C O} = n_{N_{2}}$
Partial pressure of $N_{2} = ($ Total Pressure) $\times$ (Mole fraction of $N_{2}$ )
$= (1 a t m) [\frac{n_{N_{2}}}{n_{N_{2}} + n_{C O}}]$
$= \frac{1}{2} a t m = 0 .5 a t m$

CHXI06:STATES OF MATTER

314104 Which gas cannot be collected over water?

1

{NH}_{3}

2

CO

3

N_{2}

4

H_{2}

CHXI06:STATES OF MATTER

314105 $2 .8 g o f N_{2}, 2 .8 g C O$ , and $4 .4 o f C O_{2}$ are found to exert a pressure of 700 torr. Find partial pressure of $N_{2}$ gas in the mixture.

1 280.8 torr

2 233.3 torr

3 300 torr

4 350 torr

Explanation:

$P_{N_{2}} =$ Mole fraction of $N_{2} \times P_{total}$
$= \frac{2 .8 / 28}{\frac{2 .8}{28} + \frac{2 .8}{28} + \frac{4 .4}{44}} \times 700$
$= \frac{0 .1}{0 .3} \times 700$
$= 233 .3 t o r r .$

CHXI06:STATES OF MATTER

314106 Equal weights of ${CH}_{4}$ and $H_{2}$ are mixed in a container at $25^{\circ} C$ . Fraction of total pressure exerted by methane is

1

\frac{1}{2}

2

\frac{1}{3}

3

\frac{1}{9}

4

\frac{8}{9}

Explanation:

Suppose w kg of each is taken
$n_{C H_{4}} = w / 16 a n d n_{H_{2}} = w / 2$
$p_{C H_{4}} = p_{t o t a l} \times$ mole fraction
$= p_{t o t a l} \times \frac{w / 16}{w / 2 + w / 16}$
$p_{C H_{4}} = p_{t o t a l} \times 1 / 9$
$∴$ Fraction of total pressure exerted by methane is $1 / 9$ .

CHXI06:STATES OF MATTER

314107 In a ten litre vessel, the total pressure of a gaseous mixture containing $H_{2}, N_{2}$ and ${CO}_{2}$ is 9.8 atm. The partial pressures of $H_{2}$ and $N_{2}$ are 3.7 and $4.2 atm$ respectively. Then the partial pressure of ${CO}_{2}$ is

1

1.9 atm

2

0.19 atm

3

2.4 atm

4

0.019 atm

Explanation:

$P_{t o t a l} = P (N_{2}) + P (H_{2}) + P (C O_{2})$
$9 .8 = 3 .7 + 4 .2 + P (C O_{2})$
$P (C O_{2}) = 1 .9 a t m$

CHXI06:STATES OF MATTER

314108 A gaseous mixture was prepared by taking equal moles of $CO$ and $N_{2}$ . If the total pressure of the mixture was found to be 1 atmosphere, the partial pressure of the nitrogen $(N_{2})$ in the mixture is

1

1 atm

2

0.9 atm

3

0.8 atm

4

0.5 atm

Explanation:

Given, $n_{C O} = n_{N_{2}}$
Partial pressure of $N_{2} = ($ Total Pressure) $\times$ (Mole fraction of $N_{2}$ )
$= (1 a t m) [\frac{n_{N_{2}}}{n_{N_{2}} + n_{C O}}]$
$= \frac{1}{2} a t m = 0 .5 a t m$

CHXI06:STATES OF MATTER

314104 Which gas cannot be collected over water?

1

{NH}_{3}

2

CO

3

N_{2}

4

H_{2}

CHXI06:STATES OF MATTER

314105 $2 .8 g o f N_{2}, 2 .8 g C O$ , and $4 .4 o f C O_{2}$ are found to exert a pressure of 700 torr. Find partial pressure of $N_{2}$ gas in the mixture.

1 280.8 torr

2 233.3 torr

3 300 torr

4 350 torr

Explanation:

$P_{N_{2}} =$ Mole fraction of $N_{2} \times P_{total}$
$= \frac{2 .8 / 28}{\frac{2 .8}{28} + \frac{2 .8}{28} + \frac{4 .4}{44}} \times 700$
$= \frac{0 .1}{0 .3} \times 700$
$= 233 .3 t o r r .$

CHXI06:STATES OF MATTER

314106 Equal weights of ${CH}_{4}$ and $H_{2}$ are mixed in a container at $25^{\circ} C$ . Fraction of total pressure exerted by methane is

1

\frac{1}{2}

2

\frac{1}{3}

3

\frac{1}{9}

4

\frac{8}{9}

Explanation:

Suppose w kg of each is taken
$n_{C H_{4}} = w / 16 a n d n_{H_{2}} = w / 2$
$p_{C H_{4}} = p_{t o t a l} \times$ mole fraction
$= p_{t o t a l} \times \frac{w / 16}{w / 2 + w / 16}$
$p_{C H_{4}} = p_{t o t a l} \times 1 / 9$
$∴$ Fraction of total pressure exerted by methane is $1 / 9$ .

CHXI06:STATES OF MATTER

314107 In a ten litre vessel, the total pressure of a gaseous mixture containing $H_{2}, N_{2}$ and ${CO}_{2}$ is 9.8 atm. The partial pressures of $H_{2}$ and $N_{2}$ are 3.7 and $4.2 atm$ respectively. Then the partial pressure of ${CO}_{2}$ is

1

1.9 atm

2

0.19 atm

3

2.4 atm

4

0.019 atm

Explanation:

$P_{t o t a l} = P (N_{2}) + P (H_{2}) + P (C O_{2})$
$9 .8 = 3 .7 + 4 .2 + P (C O_{2})$
$P (C O_{2}) = 1 .9 a t m$

CHXI06:STATES OF MATTER

314108 A gaseous mixture was prepared by taking equal moles of $CO$ and $N_{2}$ . If the total pressure of the mixture was found to be 1 atmosphere, the partial pressure of the nitrogen $(N_{2})$ in the mixture is

1

1 atm

2

0.9 atm

3

0.8 atm

4

0.5 atm

Explanation:

Given, $n_{C O} = n_{N_{2}}$
Partial pressure of $N_{2} = ($ Total Pressure) $\times$ (Mole fraction of $N_{2}$ )
$= (1 a t m) [\frac{n_{N_{2}}}{n_{N_{2}} + n_{C O}}]$
$= \frac{1}{2} a t m = 0 .5 a t m$

CHXI06:STATES OF MATTER

314104 Which gas cannot be collected over water?

1

{NH}_{3}

2

CO

3

N_{2}

4

H_{2}

CHXI06:STATES OF MATTER

314105 $2 .8 g o f N_{2}, 2 .8 g C O$ , and $4 .4 o f C O_{2}$ are found to exert a pressure of 700 torr. Find partial pressure of $N_{2}$ gas in the mixture.

1 280.8 torr

2 233.3 torr

3 300 torr

4 350 torr

Explanation:

$P_{N_{2}} =$ Mole fraction of $N_{2} \times P_{total}$
$= \frac{2 .8 / 28}{\frac{2 .8}{28} + \frac{2 .8}{28} + \frac{4 .4}{44}} \times 700$
$= \frac{0 .1}{0 .3} \times 700$
$= 233 .3 t o r r .$

CHXI06:STATES OF MATTER

314106 Equal weights of ${CH}_{4}$ and $H_{2}$ are mixed in a container at $25^{\circ} C$ . Fraction of total pressure exerted by methane is

1

\frac{1}{2}

2

\frac{1}{3}

3

\frac{1}{9}

4

\frac{8}{9}

Explanation:

Suppose w kg of each is taken
$n_{C H_{4}} = w / 16 a n d n_{H_{2}} = w / 2$
$p_{C H_{4}} = p_{t o t a l} \times$ mole fraction
$= p_{t o t a l} \times \frac{w / 16}{w / 2 + w / 16}$
$p_{C H_{4}} = p_{t o t a l} \times 1 / 9$
$∴$ Fraction of total pressure exerted by methane is $1 / 9$ .

CHXI06:STATES OF MATTER

314107 In a ten litre vessel, the total pressure of a gaseous mixture containing $H_{2}, N_{2}$ and ${CO}_{2}$ is 9.8 atm. The partial pressures of $H_{2}$ and $N_{2}$ are 3.7 and $4.2 atm$ respectively. Then the partial pressure of ${CO}_{2}$ is

1

1.9 atm

2

0.19 atm

3

2.4 atm

4

0.019 atm

Explanation:

$P_{t o t a l} = P (N_{2}) + P (H_{2}) + P (C O_{2})$
$9 .8 = 3 .7 + 4 .2 + P (C O_{2})$
$P (C O_{2}) = 1 .9 a t m$

CHXI06:STATES OF MATTER

314108 A gaseous mixture was prepared by taking equal moles of $CO$ and $N_{2}$ . If the total pressure of the mixture was found to be 1 atmosphere, the partial pressure of the nitrogen $(N_{2})$ in the mixture is

1

1 atm

2

0.9 atm

3

0.8 atm

4

0.5 atm

Explanation:

Given, $n_{C O} = n_{N_{2}}$
Partial pressure of $N_{2} = ($ Total Pressure) $\times$ (Mole fraction of $N_{2}$ )
$= (1 a t m) [\frac{n_{N_{2}}}{n_{N_{2}} + n_{C O}}]$
$= \frac{1}{2} a t m = 0 .5 a t m$

CHXI06:STATES OF MATTER

314104 Which gas cannot be collected over water?

1

{NH}_{3}

2

CO

3

N_{2}

4

H_{2}

CHXI06:STATES OF MATTER

314105 $2 .8 g o f N_{2}, 2 .8 g C O$ , and $4 .4 o f C O_{2}$ are found to exert a pressure of 700 torr. Find partial pressure of $N_{2}$ gas in the mixture.

1 280.8 torr

2 233.3 torr

3 300 torr

4 350 torr

Explanation:

$P_{N_{2}} =$ Mole fraction of $N_{2} \times P_{total}$
$= \frac{2 .8 / 28}{\frac{2 .8}{28} + \frac{2 .8}{28} + \frac{4 .4}{44}} \times 700$
$= \frac{0 .1}{0 .3} \times 700$
$= 233 .3 t o r r .$

CHXI06:STATES OF MATTER

314106 Equal weights of ${CH}_{4}$ and $H_{2}$ are mixed in a container at $25^{\circ} C$ . Fraction of total pressure exerted by methane is

1

\frac{1}{2}

2

\frac{1}{3}

3

\frac{1}{9}

4

\frac{8}{9}

Explanation:

Suppose w kg of each is taken
$n_{C H_{4}} = w / 16 a n d n_{H_{2}} = w / 2$
$p_{C H_{4}} = p_{t o t a l} \times$ mole fraction
$= p_{t o t a l} \times \frac{w / 16}{w / 2 + w / 16}$
$p_{C H_{4}} = p_{t o t a l} \times 1 / 9$
$∴$ Fraction of total pressure exerted by methane is $1 / 9$ .

CHXI06:STATES OF MATTER

314107 In a ten litre vessel, the total pressure of a gaseous mixture containing $H_{2}, N_{2}$ and ${CO}_{2}$ is 9.8 atm. The partial pressures of $H_{2}$ and $N_{2}$ are 3.7 and $4.2 atm$ respectively. Then the partial pressure of ${CO}_{2}$ is

1

1.9 atm

2

0.19 atm

3

2.4 atm

4

0.019 atm

Explanation:

$P_{t o t a l} = P (N_{2}) + P (H_{2}) + P (C O_{2})$
$9 .8 = 3 .7 + 4 .2 + P (C O_{2})$
$P (C O_{2}) = 1 .9 a t m$

CHXI06:STATES OF MATTER

314108 A gaseous mixture was prepared by taking equal moles of $CO$ and $N_{2}$ . If the total pressure of the mixture was found to be 1 atmosphere, the partial pressure of the nitrogen $(N_{2})$ in the mixture is

1

1 atm

2

0.9 atm

3

0.8 atm

4

0.5 atm

Explanation:

Given, $n_{C O} = n_{N_{2}}$
Partial pressure of $N_{2} = ($ Total Pressure) $\times$ (Mole fraction of $N_{2}$ )
$= (1 a t m) [\frac{n_{N_{2}}}{n_{N_{2}} + n_{C O}}]$
$= \frac{1}{2} a t m = 0 .5 a t m$

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