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CHXI06:STATES OF MATTER

314176 A mixture of $C_{2} H_{6}$ and $C_{2} H_{4}$ occupies 40 litres at $1 atm$ and at $400 K$ . The mixture reacts completely with $130 g$ of $O_{2}$ to produce ${CO}_{2}$ and $H_{2} O$ . Assuming ideal gas behaviour, the mole fractions of $C_{2} H_{4}$ in the mixture is

1 0.33

2 0.67

3 0.43

4 0.57

Explanation:

Let the number of moles of $C_{2} H_{6}$ and $C_{2} H_{4}$ be $n_{1}$ and $n_{2}$ respectively. Applying ideal gas equation, $P V = n R T$ .
$1 \times 10 = (n_{1} + n_{2}) \times 0 .0821 \times 400 (1)$
$C_{2} H_{6} + C_{2} H_{4} + O_{2} \to {CO}_{2} + H_{2} O$
For $C, 2 n_{1} + 2 n_{2} = M o l e s o f C O_{2} (2)$
For $H, 6 n_{1} + 4 n_{2} = 2 \times M o l e s o f H_{2} O (3)$
For $O, 2 \times \frac{130}{32} = 2 \times M o l e s o f C O_{2} + M o l e s o f H_{2} O (4)$
From equations (2), (3) and (4) we get,
$7 n_{1} + 6 n_{2} = \frac{260}{32}$
Solving equations (1) and (4) we get,
$n_{1} = 0.8168; n_{2} = 0.4012$
$∴$ Mole of fraction of
$C_{2} H_{6} = \frac{0.8168}{0.8168 + 0.4012} = 0.67$
Mole fraction $C_{2} H_{4} = 1 - 0.67 = 0.33$

CHXI06:STATES OF MATTER

314177 8.2L of an ideal gas weighs $9.0 g$ at $300 K$ and $1 atm$ pressure. The molecular mass of the gas is

1 9

2 27

3 54

4 81

Explanation:

Given:
$V = 8.2 L, w = 9.0 g, T = 300 K$
Ideal gas equation is pV=nRT or pV = $\frac{w}{M} R T$
$M = \frac{w R T}{p V} = \frac{9.0 \times 0.082 \times 300}{1 \times 8.2}$
$= 27 g m o l^{- 1}$

CHXI06:STATES OF MATTER

314178 The value of the universal gas constant $(R)$ depends upon the

1 Nature of the gas

2 Mass of the gas

3 Temperature of the gas

4 Units of measurement

Explanation:

The value of the universal gas constant, $R$ depends upon the units of measurement.
$R = 0.0821 L atm {mol}^{- 1} K^{- 1}$
$= 8.314 J {mol}^{- 1} K^{- 1}$
$= 1.98 cal {mol}^{- 1} K^{- 1}$

CHXI06:STATES OF MATTER

314179 Choose the correct option for the total pressure (in atm.) in a mixture of $4 g O_{2}$ and $2 g H_{2}$ confined in a total volume of one litre at $0^{\circ} C$ is : [Given $R = 0.082 L atm {mol}^{- 1} K^{- 1}, T = 273 K$ ]

1 2.602

2 25.18

3 26.02

4 2.518

Explanation:

No. of moles of $H_{2} = \frac{2}{2} = 1$ mole
No. of moles of $O_{2} = \frac{4}{32} = \frac{1}{8}$ mole
Total no. of moles $= 1 + \frac{1}{8} = \frac{9}{8}$ mole
According to Ideal gas equation, $P V = n R T$
$P = \frac{n R T}{V} = \frac{9}{8} \times \frac{0.082 \times 273}{1}$
$P = 25.18 atm$

NEET - 2021

CHXI06:STATES OF MATTER

314176 A mixture of $C_{2} H_{6}$ and $C_{2} H_{4}$ occupies 40 litres at $1 atm$ and at $400 K$ . The mixture reacts completely with $130 g$ of $O_{2}$ to produce ${CO}_{2}$ and $H_{2} O$ . Assuming ideal gas behaviour, the mole fractions of $C_{2} H_{4}$ in the mixture is

1 0.33

2 0.67

3 0.43

4 0.57

Explanation:

Let the number of moles of $C_{2} H_{6}$ and $C_{2} H_{4}$ be $n_{1}$ and $n_{2}$ respectively. Applying ideal gas equation, $P V = n R T$ .
$1 \times 10 = (n_{1} + n_{2}) \times 0 .0821 \times 400 (1)$
$C_{2} H_{6} + C_{2} H_{4} + O_{2} \to {CO}_{2} + H_{2} O$
For $C, 2 n_{1} + 2 n_{2} = M o l e s o f C O_{2} (2)$
For $H, 6 n_{1} + 4 n_{2} = 2 \times M o l e s o f H_{2} O (3)$
For $O, 2 \times \frac{130}{32} = 2 \times M o l e s o f C O_{2} + M o l e s o f H_{2} O (4)$
From equations (2), (3) and (4) we get,
$7 n_{1} + 6 n_{2} = \frac{260}{32}$
Solving equations (1) and (4) we get,
$n_{1} = 0.8168; n_{2} = 0.4012$
$∴$ Mole of fraction of
$C_{2} H_{6} = \frac{0.8168}{0.8168 + 0.4012} = 0.67$
Mole fraction $C_{2} H_{4} = 1 - 0.67 = 0.33$

CHXI06:STATES OF MATTER

314177 8.2L of an ideal gas weighs $9.0 g$ at $300 K$ and $1 atm$ pressure. The molecular mass of the gas is

1 9

2 27

3 54

4 81

Explanation:

Given:
$V = 8.2 L, w = 9.0 g, T = 300 K$
Ideal gas equation is pV=nRT or pV = $\frac{w}{M} R T$
$M = \frac{w R T}{p V} = \frac{9.0 \times 0.082 \times 300}{1 \times 8.2}$
$= 27 g m o l^{- 1}$

CHXI06:STATES OF MATTER

314178 The value of the universal gas constant $(R)$ depends upon the

1 Nature of the gas

2 Mass of the gas

3 Temperature of the gas

4 Units of measurement

Explanation:

The value of the universal gas constant, $R$ depends upon the units of measurement.
$R = 0.0821 L atm {mol}^{- 1} K^{- 1}$
$= 8.314 J {mol}^{- 1} K^{- 1}$
$= 1.98 cal {mol}^{- 1} K^{- 1}$

CHXI06:STATES OF MATTER

314179 Choose the correct option for the total pressure (in atm.) in a mixture of $4 g O_{2}$ and $2 g H_{2}$ confined in a total volume of one litre at $0^{\circ} C$ is : [Given $R = 0.082 L atm {mol}^{- 1} K^{- 1}, T = 273 K$ ]

1 2.602

2 25.18

3 26.02

4 2.518

Explanation:

No. of moles of $H_{2} = \frac{2}{2} = 1$ mole
No. of moles of $O_{2} = \frac{4}{32} = \frac{1}{8}$ mole
Total no. of moles $= 1 + \frac{1}{8} = \frac{9}{8}$ mole
According to Ideal gas equation, $P V = n R T$
$P = \frac{n R T}{V} = \frac{9}{8} \times \frac{0.082 \times 273}{1}$
$P = 25.18 atm$

NEET - 2021

CHXI06:STATES OF MATTER

314176 A mixture of $C_{2} H_{6}$ and $C_{2} H_{4}$ occupies 40 litres at $1 atm$ and at $400 K$ . The mixture reacts completely with $130 g$ of $O_{2}$ to produce ${CO}_{2}$ and $H_{2} O$ . Assuming ideal gas behaviour, the mole fractions of $C_{2} H_{4}$ in the mixture is

1 0.33

2 0.67

3 0.43

4 0.57

Explanation:

Let the number of moles of $C_{2} H_{6}$ and $C_{2} H_{4}$ be $n_{1}$ and $n_{2}$ respectively. Applying ideal gas equation, $P V = n R T$ .
$1 \times 10 = (n_{1} + n_{2}) \times 0 .0821 \times 400 (1)$
$C_{2} H_{6} + C_{2} H_{4} + O_{2} \to {CO}_{2} + H_{2} O$
For $C, 2 n_{1} + 2 n_{2} = M o l e s o f C O_{2} (2)$
For $H, 6 n_{1} + 4 n_{2} = 2 \times M o l e s o f H_{2} O (3)$
For $O, 2 \times \frac{130}{32} = 2 \times M o l e s o f C O_{2} + M o l e s o f H_{2} O (4)$
From equations (2), (3) and (4) we get,
$7 n_{1} + 6 n_{2} = \frac{260}{32}$
Solving equations (1) and (4) we get,
$n_{1} = 0.8168; n_{2} = 0.4012$
$∴$ Mole of fraction of
$C_{2} H_{6} = \frac{0.8168}{0.8168 + 0.4012} = 0.67$
Mole fraction $C_{2} H_{4} = 1 - 0.67 = 0.33$

CHXI06:STATES OF MATTER

314177 8.2L of an ideal gas weighs $9.0 g$ at $300 K$ and $1 atm$ pressure. The molecular mass of the gas is

1 9

2 27

3 54

4 81

Explanation:

Given:
$V = 8.2 L, w = 9.0 g, T = 300 K$
Ideal gas equation is pV=nRT or pV = $\frac{w}{M} R T$
$M = \frac{w R T}{p V} = \frac{9.0 \times 0.082 \times 300}{1 \times 8.2}$
$= 27 g m o l^{- 1}$

CHXI06:STATES OF MATTER

314178 The value of the universal gas constant $(R)$ depends upon the

1 Nature of the gas

2 Mass of the gas

3 Temperature of the gas

4 Units of measurement

Explanation:

The value of the universal gas constant, $R$ depends upon the units of measurement.
$R = 0.0821 L atm {mol}^{- 1} K^{- 1}$
$= 8.314 J {mol}^{- 1} K^{- 1}$
$= 1.98 cal {mol}^{- 1} K^{- 1}$

CHXI06:STATES OF MATTER

314179 Choose the correct option for the total pressure (in atm.) in a mixture of $4 g O_{2}$ and $2 g H_{2}$ confined in a total volume of one litre at $0^{\circ} C$ is : [Given $R = 0.082 L atm {mol}^{- 1} K^{- 1}, T = 273 K$ ]

1 2.602

2 25.18

3 26.02

4 2.518

Explanation:

No. of moles of $H_{2} = \frac{2}{2} = 1$ mole
No. of moles of $O_{2} = \frac{4}{32} = \frac{1}{8}$ mole
Total no. of moles $= 1 + \frac{1}{8} = \frac{9}{8}$ mole
According to Ideal gas equation, $P V = n R T$
$P = \frac{n R T}{V} = \frac{9}{8} \times \frac{0.082 \times 273}{1}$
$P = 25.18 atm$

NEET - 2021

CHXI06:STATES OF MATTER

314176 A mixture of $C_{2} H_{6}$ and $C_{2} H_{4}$ occupies 40 litres at $1 atm$ and at $400 K$ . The mixture reacts completely with $130 g$ of $O_{2}$ to produce ${CO}_{2}$ and $H_{2} O$ . Assuming ideal gas behaviour, the mole fractions of $C_{2} H_{4}$ in the mixture is

1 0.33

2 0.67

3 0.43

4 0.57

Explanation:

Let the number of moles of $C_{2} H_{6}$ and $C_{2} H_{4}$ be $n_{1}$ and $n_{2}$ respectively. Applying ideal gas equation, $P V = n R T$ .
$1 \times 10 = (n_{1} + n_{2}) \times 0 .0821 \times 400 (1)$
$C_{2} H_{6} + C_{2} H_{4} + O_{2} \to {CO}_{2} + H_{2} O$
For $C, 2 n_{1} + 2 n_{2} = M o l e s o f C O_{2} (2)$
For $H, 6 n_{1} + 4 n_{2} = 2 \times M o l e s o f H_{2} O (3)$
For $O, 2 \times \frac{130}{32} = 2 \times M o l e s o f C O_{2} + M o l e s o f H_{2} O (4)$
From equations (2), (3) and (4) we get,
$7 n_{1} + 6 n_{2} = \frac{260}{32}$
Solving equations (1) and (4) we get,
$n_{1} = 0.8168; n_{2} = 0.4012$
$∴$ Mole of fraction of
$C_{2} H_{6} = \frac{0.8168}{0.8168 + 0.4012} = 0.67$
Mole fraction $C_{2} H_{4} = 1 - 0.67 = 0.33$

CHXI06:STATES OF MATTER

314177 8.2L of an ideal gas weighs $9.0 g$ at $300 K$ and $1 atm$ pressure. The molecular mass of the gas is

1 9

2 27

3 54

4 81

Explanation:

Given:
$V = 8.2 L, w = 9.0 g, T = 300 K$
Ideal gas equation is pV=nRT or pV = $\frac{w}{M} R T$
$M = \frac{w R T}{p V} = \frac{9.0 \times 0.082 \times 300}{1 \times 8.2}$
$= 27 g m o l^{- 1}$

CHXI06:STATES OF MATTER

314178 The value of the universal gas constant $(R)$ depends upon the

1 Nature of the gas

2 Mass of the gas

3 Temperature of the gas

4 Units of measurement

Explanation:

The value of the universal gas constant, $R$ depends upon the units of measurement.
$R = 0.0821 L atm {mol}^{- 1} K^{- 1}$
$= 8.314 J {mol}^{- 1} K^{- 1}$
$= 1.98 cal {mol}^{- 1} K^{- 1}$

CHXI06:STATES OF MATTER

314179 Choose the correct option for the total pressure (in atm.) in a mixture of $4 g O_{2}$ and $2 g H_{2}$ confined in a total volume of one litre at $0^{\circ} C$ is : [Given $R = 0.082 L atm {mol}^{- 1} K^{- 1}, T = 273 K$ ]

1 2.602

2 25.18

3 26.02

4 2.518

Explanation:

No. of moles of $H_{2} = \frac{2}{2} = 1$ mole
No. of moles of $O_{2} = \frac{4}{32} = \frac{1}{8}$ mole
Total no. of moles $= 1 + \frac{1}{8} = \frac{9}{8}$ mole
According to Ideal gas equation, $P V = n R T$
$P = \frac{n R T}{V} = \frac{9}{8} \times \frac{0.082 \times 273}{1}$
$P = 25.18 atm$

NEET - 2021

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