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CHXI06:STATES OF MATTER

314189 At 1 atm pressure, 0.5 g of a gas has the volume of 0.1 L at 300 K . Gas constant $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ . Find the molar mass of the gas.

1 112

2 24.6

3 123.1

4 49.2

Explanation:

$PV = n RT$
$\Rightarrow 1 \times 0.1 = \frac{0.5}{M} \times 0.0821 \times 300$
$\Rightarrow M = \frac{0.5 \times 0.0821 \times 300}{0.1} = 123.1 g m o l^{- 1}$

CHXI06:STATES OF MATTER

314190 A bubble of air is under water at temperature $15^{\circ} C$ and the pressure 1.5 bar. If the bubble rises to the surface where the temperature is $25^{\circ} C$ and pressure $1.0 bar$ , then the volume of bubble

1 will become greater by a factor of 1.6

2 will become greater by a factor of 1.1

3 will become smaller by a factor of 0.70

4 will become greater by a factor of 2.5

Explanation:

$\frac{P_{2} V_{2}}{T_{2}} = \frac{P_{1} V_{1}}{T_{1}}$
$V_{2} = \frac{1.5 \times V_{1}}{288} \times \frac{298}{1} = 1.55 V_{1}$

CHXI06:STATES OF MATTER

314191 What will be the pressure (in atm) exerted by a mixture of $3.2 g$ of methane and $4.4 g$ of carbon dioxide contained in a $9 d m^{3}$ flask at $27^{\circ} C$ ?

1 0.0821

2 0.00821

3 8.21

4 0.821

Explanation:

Number of moles of methane $\frac{3.2}{16} = 0.2$
Number of moles of ${CO}_{2} = \frac{4.4}{44} = 0.1$
Total number of moles $= 0.3$
$PV = nRT$
$P \times 9 = 0.3 \times 0.0821 \times 300$
$P = 0.821 atm$

CHXI06:STATES OF MATTER

314192 A compound exists in the gaseous phase both as monomer (A) and dimer $(A_{2})$ . The molecular weight of $A$ is 48 . In an experiment $96 g$ of the compound was confined in a vessel of volume 33.6 litre and heated to $273^{\circ} C$ . The pressure developed if the compound exists as dimer to the extent of $50 %$ by weight under these conditions is

1

2 atm

2

4 atm

3

3 atm

4

5 atm

Explanation:

Since, $A$ and $A_{2}$ are two states in gaseous phase having their weight ratio $50 %$ , i.e., $1 : 1$
Moles of $A = \frac{96}{2} \times \frac{1}{48} = 1$ ;
Moles of $A_{2} = \frac{96}{2} \times \frac{1}{96} = \frac{1}{2}$
Total moles of A and $A_{2}$ are $= 1 + \frac{1}{2} = \frac{3}{2}$
$P V = n R T$
$P \times 33.6 = \frac{3}{2} \times 0.0821 \times 546$
$P = 2 atm$

CHXI06:STATES OF MATTER

314193 Which of the following indicates correct $R$ (Gas constant) values in different units?
(I) 1.987 cals ${mole}^{- 1} K^{- 1}$
(II) $8.314 \times 10^{7} J {mole}^{- 1} K^{- 1}$
(III) 0.0821 lit-atm mole $K^{- 1}$
(IV) 82.1 lit. ${mole}^{- 1} K^{- 1}$

1 I, II

2 II, III

3 I, III

4 IV, III

Explanation:

$R = 8.314 J {mol}^{- 1} K^{- 1}$
$= 82.1 mL {mol}^{- 1} K^{- 1}$
$= 0 .0821 L a t m m o l^{- 1} K^{- 1}$
$= 1 .987 c a l m o l^{- 1} K^{- 1}$

CHXI06:STATES OF MATTER

314189 At 1 atm pressure, 0.5 g of a gas has the volume of 0.1 L at 300 K . Gas constant $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ . Find the molar mass of the gas.

1 112

2 24.6

3 123.1

4 49.2

Explanation:

$PV = n RT$
$\Rightarrow 1 \times 0.1 = \frac{0.5}{M} \times 0.0821 \times 300$
$\Rightarrow M = \frac{0.5 \times 0.0821 \times 300}{0.1} = 123.1 g m o l^{- 1}$

CHXI06:STATES OF MATTER

314190 A bubble of air is under water at temperature $15^{\circ} C$ and the pressure 1.5 bar. If the bubble rises to the surface where the temperature is $25^{\circ} C$ and pressure $1.0 bar$ , then the volume of bubble

1 will become greater by a factor of 1.6

2 will become greater by a factor of 1.1

3 will become smaller by a factor of 0.70

4 will become greater by a factor of 2.5

Explanation:

$\frac{P_{2} V_{2}}{T_{2}} = \frac{P_{1} V_{1}}{T_{1}}$
$V_{2} = \frac{1.5 \times V_{1}}{288} \times \frac{298}{1} = 1.55 V_{1}$

CHXI06:STATES OF MATTER

314191 What will be the pressure (in atm) exerted by a mixture of $3.2 g$ of methane and $4.4 g$ of carbon dioxide contained in a $9 d m^{3}$ flask at $27^{\circ} C$ ?

1 0.0821

2 0.00821

3 8.21

4 0.821

Explanation:

Number of moles of methane $\frac{3.2}{16} = 0.2$
Number of moles of ${CO}_{2} = \frac{4.4}{44} = 0.1$
Total number of moles $= 0.3$
$PV = nRT$
$P \times 9 = 0.3 \times 0.0821 \times 300$
$P = 0.821 atm$

CHXI06:STATES OF MATTER

314192 A compound exists in the gaseous phase both as monomer (A) and dimer $(A_{2})$ . The molecular weight of $A$ is 48 . In an experiment $96 g$ of the compound was confined in a vessel of volume 33.6 litre and heated to $273^{\circ} C$ . The pressure developed if the compound exists as dimer to the extent of $50 %$ by weight under these conditions is

1

2 atm

2

4 atm

3

3 atm

4

5 atm

Explanation:

Since, $A$ and $A_{2}$ are two states in gaseous phase having their weight ratio $50 %$ , i.e., $1 : 1$
Moles of $A = \frac{96}{2} \times \frac{1}{48} = 1$ ;
Moles of $A_{2} = \frac{96}{2} \times \frac{1}{96} = \frac{1}{2}$
Total moles of A and $A_{2}$ are $= 1 + \frac{1}{2} = \frac{3}{2}$
$P V = n R T$
$P \times 33.6 = \frac{3}{2} \times 0.0821 \times 546$
$P = 2 atm$

CHXI06:STATES OF MATTER

314193 Which of the following indicates correct $R$ (Gas constant) values in different units?
(I) 1.987 cals ${mole}^{- 1} K^{- 1}$
(II) $8.314 \times 10^{7} J {mole}^{- 1} K^{- 1}$
(III) 0.0821 lit-atm mole $K^{- 1}$
(IV) 82.1 lit. ${mole}^{- 1} K^{- 1}$

1 I, II

2 II, III

3 I, III

4 IV, III

Explanation:

$R = 8.314 J {mol}^{- 1} K^{- 1}$
$= 82.1 mL {mol}^{- 1} K^{- 1}$
$= 0 .0821 L a t m m o l^{- 1} K^{- 1}$
$= 1 .987 c a l m o l^{- 1} K^{- 1}$

CHXI06:STATES OF MATTER

314189 At 1 atm pressure, 0.5 g of a gas has the volume of 0.1 L at 300 K . Gas constant $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ . Find the molar mass of the gas.

1 112

2 24.6

3 123.1

4 49.2

Explanation:

$PV = n RT$
$\Rightarrow 1 \times 0.1 = \frac{0.5}{M} \times 0.0821 \times 300$
$\Rightarrow M = \frac{0.5 \times 0.0821 \times 300}{0.1} = 123.1 g m o l^{- 1}$

CHXI06:STATES OF MATTER

314190 A bubble of air is under water at temperature $15^{\circ} C$ and the pressure 1.5 bar. If the bubble rises to the surface where the temperature is $25^{\circ} C$ and pressure $1.0 bar$ , then the volume of bubble

1 will become greater by a factor of 1.6

2 will become greater by a factor of 1.1

3 will become smaller by a factor of 0.70

4 will become greater by a factor of 2.5

Explanation:

$\frac{P_{2} V_{2}}{T_{2}} = \frac{P_{1} V_{1}}{T_{1}}$
$V_{2} = \frac{1.5 \times V_{1}}{288} \times \frac{298}{1} = 1.55 V_{1}$

CHXI06:STATES OF MATTER

314191 What will be the pressure (in atm) exerted by a mixture of $3.2 g$ of methane and $4.4 g$ of carbon dioxide contained in a $9 d m^{3}$ flask at $27^{\circ} C$ ?

1 0.0821

2 0.00821

3 8.21

4 0.821

Explanation:

Number of moles of methane $\frac{3.2}{16} = 0.2$
Number of moles of ${CO}_{2} = \frac{4.4}{44} = 0.1$
Total number of moles $= 0.3$
$PV = nRT$
$P \times 9 = 0.3 \times 0.0821 \times 300$
$P = 0.821 atm$

CHXI06:STATES OF MATTER

314192 A compound exists in the gaseous phase both as monomer (A) and dimer $(A_{2})$ . The molecular weight of $A$ is 48 . In an experiment $96 g$ of the compound was confined in a vessel of volume 33.6 litre and heated to $273^{\circ} C$ . The pressure developed if the compound exists as dimer to the extent of $50 %$ by weight under these conditions is

1

2 atm

2

4 atm

3

3 atm

4

5 atm

Explanation:

Since, $A$ and $A_{2}$ are two states in gaseous phase having their weight ratio $50 %$ , i.e., $1 : 1$
Moles of $A = \frac{96}{2} \times \frac{1}{48} = 1$ ;
Moles of $A_{2} = \frac{96}{2} \times \frac{1}{96} = \frac{1}{2}$
Total moles of A and $A_{2}$ are $= 1 + \frac{1}{2} = \frac{3}{2}$
$P V = n R T$
$P \times 33.6 = \frac{3}{2} \times 0.0821 \times 546$
$P = 2 atm$

CHXI06:STATES OF MATTER

314193 Which of the following indicates correct $R$ (Gas constant) values in different units?
(I) 1.987 cals ${mole}^{- 1} K^{- 1}$
(II) $8.314 \times 10^{7} J {mole}^{- 1} K^{- 1}$
(III) 0.0821 lit-atm mole $K^{- 1}$
(IV) 82.1 lit. ${mole}^{- 1} K^{- 1}$

1 I, II

2 II, III

3 I, III

4 IV, III

Explanation:

$R = 8.314 J {mol}^{- 1} K^{- 1}$
$= 82.1 mL {mol}^{- 1} K^{- 1}$
$= 0 .0821 L a t m m o l^{- 1} K^{- 1}$
$= 1 .987 c a l m o l^{- 1} K^{- 1}$

CHXI06:STATES OF MATTER

314189 At 1 atm pressure, 0.5 g of a gas has the volume of 0.1 L at 300 K . Gas constant $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ . Find the molar mass of the gas.

1 112

2 24.6

3 123.1

4 49.2

Explanation:

$PV = n RT$
$\Rightarrow 1 \times 0.1 = \frac{0.5}{M} \times 0.0821 \times 300$
$\Rightarrow M = \frac{0.5 \times 0.0821 \times 300}{0.1} = 123.1 g m o l^{- 1}$

CHXI06:STATES OF MATTER

314190 A bubble of air is under water at temperature $15^{\circ} C$ and the pressure 1.5 bar. If the bubble rises to the surface where the temperature is $25^{\circ} C$ and pressure $1.0 bar$ , then the volume of bubble

1 will become greater by a factor of 1.6

2 will become greater by a factor of 1.1

3 will become smaller by a factor of 0.70

4 will become greater by a factor of 2.5

Explanation:

$\frac{P_{2} V_{2}}{T_{2}} = \frac{P_{1} V_{1}}{T_{1}}$
$V_{2} = \frac{1.5 \times V_{1}}{288} \times \frac{298}{1} = 1.55 V_{1}$

CHXI06:STATES OF MATTER

314191 What will be the pressure (in atm) exerted by a mixture of $3.2 g$ of methane and $4.4 g$ of carbon dioxide contained in a $9 d m^{3}$ flask at $27^{\circ} C$ ?

1 0.0821

2 0.00821

3 8.21

4 0.821

Explanation:

Number of moles of methane $\frac{3.2}{16} = 0.2$
Number of moles of ${CO}_{2} = \frac{4.4}{44} = 0.1$
Total number of moles $= 0.3$
$PV = nRT$
$P \times 9 = 0.3 \times 0.0821 \times 300$
$P = 0.821 atm$

CHXI06:STATES OF MATTER

314192 A compound exists in the gaseous phase both as monomer (A) and dimer $(A_{2})$ . The molecular weight of $A$ is 48 . In an experiment $96 g$ of the compound was confined in a vessel of volume 33.6 litre and heated to $273^{\circ} C$ . The pressure developed if the compound exists as dimer to the extent of $50 %$ by weight under these conditions is

1

2 atm

2

4 atm

3

3 atm

4

5 atm

Explanation:

Since, $A$ and $A_{2}$ are two states in gaseous phase having their weight ratio $50 %$ , i.e., $1 : 1$
Moles of $A = \frac{96}{2} \times \frac{1}{48} = 1$ ;
Moles of $A_{2} = \frac{96}{2} \times \frac{1}{96} = \frac{1}{2}$
Total moles of A and $A_{2}$ are $= 1 + \frac{1}{2} = \frac{3}{2}$
$P V = n R T$
$P \times 33.6 = \frac{3}{2} \times 0.0821 \times 546$
$P = 2 atm$

CHXI06:STATES OF MATTER

314193 Which of the following indicates correct $R$ (Gas constant) values in different units?
(I) 1.987 cals ${mole}^{- 1} K^{- 1}$
(II) $8.314 \times 10^{7} J {mole}^{- 1} K^{- 1}$
(III) 0.0821 lit-atm mole $K^{- 1}$
(IV) 82.1 lit. ${mole}^{- 1} K^{- 1}$

1 I, II

2 II, III

3 I, III

4 IV, III

Explanation:

$R = 8.314 J {mol}^{- 1} K^{- 1}$
$= 82.1 mL {mol}^{- 1} K^{- 1}$
$= 0 .0821 L a t m m o l^{- 1} K^{- 1}$
$= 1 .987 c a l m o l^{- 1} K^{- 1}$

CHXI06:STATES OF MATTER

314189 At 1 atm pressure, 0.5 g of a gas has the volume of 0.1 L at 300 K . Gas constant $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ . Find the molar mass of the gas.

1 112

2 24.6

3 123.1

4 49.2

Explanation:

$PV = n RT$
$\Rightarrow 1 \times 0.1 = \frac{0.5}{M} \times 0.0821 \times 300$
$\Rightarrow M = \frac{0.5 \times 0.0821 \times 300}{0.1} = 123.1 g m o l^{- 1}$

CHXI06:STATES OF MATTER

314190 A bubble of air is under water at temperature $15^{\circ} C$ and the pressure 1.5 bar. If the bubble rises to the surface where the temperature is $25^{\circ} C$ and pressure $1.0 bar$ , then the volume of bubble

1 will become greater by a factor of 1.6

2 will become greater by a factor of 1.1

3 will become smaller by a factor of 0.70

4 will become greater by a factor of 2.5

Explanation:

$\frac{P_{2} V_{2}}{T_{2}} = \frac{P_{1} V_{1}}{T_{1}}$
$V_{2} = \frac{1.5 \times V_{1}}{288} \times \frac{298}{1} = 1.55 V_{1}$

CHXI06:STATES OF MATTER

314191 What will be the pressure (in atm) exerted by a mixture of $3.2 g$ of methane and $4.4 g$ of carbon dioxide contained in a $9 d m^{3}$ flask at $27^{\circ} C$ ?

1 0.0821

2 0.00821

3 8.21

4 0.821

Explanation:

Number of moles of methane $\frac{3.2}{16} = 0.2$
Number of moles of ${CO}_{2} = \frac{4.4}{44} = 0.1$
Total number of moles $= 0.3$
$PV = nRT$
$P \times 9 = 0.3 \times 0.0821 \times 300$
$P = 0.821 atm$

CHXI06:STATES OF MATTER

314192 A compound exists in the gaseous phase both as monomer (A) and dimer $(A_{2})$ . The molecular weight of $A$ is 48 . In an experiment $96 g$ of the compound was confined in a vessel of volume 33.6 litre and heated to $273^{\circ} C$ . The pressure developed if the compound exists as dimer to the extent of $50 %$ by weight under these conditions is

1

2 atm

2

4 atm

3

3 atm

4

5 atm

Explanation:

Since, $A$ and $A_{2}$ are two states in gaseous phase having their weight ratio $50 %$ , i.e., $1 : 1$
Moles of $A = \frac{96}{2} \times \frac{1}{48} = 1$ ;
Moles of $A_{2} = \frac{96}{2} \times \frac{1}{96} = \frac{1}{2}$
Total moles of A and $A_{2}$ are $= 1 + \frac{1}{2} = \frac{3}{2}$
$P V = n R T$
$P \times 33.6 = \frac{3}{2} \times 0.0821 \times 546$
$P = 2 atm$

CHXI06:STATES OF MATTER

314193 Which of the following indicates correct $R$ (Gas constant) values in different units?
(I) 1.987 cals ${mole}^{- 1} K^{- 1}$
(II) $8.314 \times 10^{7} J {mole}^{- 1} K^{- 1}$
(III) 0.0821 lit-atm mole $K^{- 1}$
(IV) 82.1 lit. ${mole}^{- 1} K^{- 1}$

1 I, II

2 II, III

3 I, III

4 IV, III

Explanation:

$R = 8.314 J {mol}^{- 1} K^{- 1}$
$= 82.1 mL {mol}^{- 1} K^{- 1}$
$= 0 .0821 L a t m m o l^{- 1} K^{- 1}$
$= 1 .987 c a l m o l^{- 1} K^{- 1}$