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CHXI06:STATES OF MATTER

314211 The volume of $0.0168 mol$ of $O_{2}$ obtained by decomposition of ${KClO}_{3}$ and collected by displacement of water is $428 mL$ at pressure $754 mm$ of $Hg$ at $25^{\circ} C$ . The pressure of water vapour (in $mm$ of $Hg$ ) at $25^{\circ} C$ is

1 18.5

2 20.6

3 22.3

4 24.6

Explanation:

Volume of $0.0168 mol$ of $O_{2}$ at $STP$
$= 0.0168 \times 22400 mL = 376.3 mL$
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} \Rightarrow \frac{760 \times 376.3}{273} = \frac{P_{2} \times 428}{298}$
$\Rightarrow P_{2} = 729.4 mm$ of $Hg$
Pressure of water vapor $= 754 - 729.4 = 24.6$ $mm$ of $Hg$

CHXI06:STATES OF MATTER

314212 If $1 0^{- 4} d m^{3}$ of water is introduced into a $1 .0 d m^{3}$ flask at $300 K$ , then the number of moles of water formed in the vapour phase when equilibrium is established is [Given: Vapour pressure of $H_{2} O$ at $300 K$ is $3170 Pa$ ; $R = 8.314 {JK}^{- 1} {mol}^{- 1}]$

1

1.27 \times 10^{- 3}

2

5.56 \times 10^{- 3}

3

1.53 \times 10^{- 2}

4

4.46 \times 10^{- 2}

Explanation:

$P V = n R T; P = 3170 Pa; V = 1 {dm}^{3} = 10^{- 3} m^{3}$
$3170 \times 10^{- 3} \times 1 = n \times 8.314 \times 300$
$n = \frac{3170 \times 10^{- 3}}{8.314 \times 300} = 1.27 \times 10^{- 3}$

CHXI06:STATES OF MATTER

314213 What will be the pressure of the gaseous mixture when $0.5 L$ of $H_{2}$ at 0.8 bar and $2 L$ of oxygen at 0.7 bar are introduced in $1 L$ vessel at $27^{\circ} C$ ?

1

1.8 bar

2

0.4 bar

3

1.4 bar

4

2.5 bar

Explanation:

Ideal gas equation is $PV = nRT$
Applying ideal gas equation for $H_{2}$ gas,
$0.8 \times 0.5 = n_{H_{2}} R T \Rightarrow n_{H_{2}} = \frac{0.8 \times 0.5}{R T} = \frac{0.4}{R T}$
Applying ideal gas equation for $O_{2}$ gas,
$0.7 \times 2.0 = n_{O_{2}} R T \Rightarrow n_{O_{2}} = \frac{0.7 \times 2.0}{R T} = \frac{1.4}{R T}$
When gas mixture is introduced in $1 L$ vessel, then
$P V = (n_{H_{2}} + n_{O_{2}}) R T$
$P \times 1 = (\frac{0 .4}{R T} + \frac{1 .4}{R T}) R T \Rightarrow P = 0 .4 + 1 .4 = 1 .8 b a r$

CHXI06:STATES OF MATTER

314214 0.5 moles of gas $A$ and $x$ moles of gas $B$ exert a pressure of $200 Pa$ in a container of volume 10 $m^{3}$ at $1000 K$ . Given $R$ is the gas constant in $J K^{- 1} m o l^{- 1} .$ The value of 'x' is

1

\frac{2 R}{4 + R}

2

\frac{2 R}{4 - R}

3

\frac{4 - R}{2 R}

4

\frac{4 + R}{2 R}

Explanation:

Total number of moles $(n_{T}) = (0 .5 + x)$
$PV = nRT$
$200 \times 10 = (0 .5 + x) R \times 1000 \frac{2}{R}$ $= 0 .5 + x$
$\Rightarrow \frac{2}{R} = \frac{1}{2} + x$
$\Rightarrow \frac{2}{R} - \frac{1}{2} = x \Rightarrow x = \frac{4 - R}{2 R}$

JEE - 2018

CHXI06:STATES OF MATTER

314211 The volume of $0.0168 mol$ of $O_{2}$ obtained by decomposition of ${KClO}_{3}$ and collected by displacement of water is $428 mL$ at pressure $754 mm$ of $Hg$ at $25^{\circ} C$ . The pressure of water vapour (in $mm$ of $Hg$ ) at $25^{\circ} C$ is

1 18.5

2 20.6

3 22.3

4 24.6

Explanation:

Volume of $0.0168 mol$ of $O_{2}$ at $STP$
$= 0.0168 \times 22400 mL = 376.3 mL$
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} \Rightarrow \frac{760 \times 376.3}{273} = \frac{P_{2} \times 428}{298}$
$\Rightarrow P_{2} = 729.4 mm$ of $Hg$
Pressure of water vapor $= 754 - 729.4 = 24.6$ $mm$ of $Hg$

CHXI06:STATES OF MATTER

314212 If $1 0^{- 4} d m^{3}$ of water is introduced into a $1 .0 d m^{3}$ flask at $300 K$ , then the number of moles of water formed in the vapour phase when equilibrium is established is [Given: Vapour pressure of $H_{2} O$ at $300 K$ is $3170 Pa$ ; $R = 8.314 {JK}^{- 1} {mol}^{- 1}]$

1

1.27 \times 10^{- 3}

2

5.56 \times 10^{- 3}

3

1.53 \times 10^{- 2}

4

4.46 \times 10^{- 2}

Explanation:

$P V = n R T; P = 3170 Pa; V = 1 {dm}^{3} = 10^{- 3} m^{3}$
$3170 \times 10^{- 3} \times 1 = n \times 8.314 \times 300$
$n = \frac{3170 \times 10^{- 3}}{8.314 \times 300} = 1.27 \times 10^{- 3}$

CHXI06:STATES OF MATTER

314213 What will be the pressure of the gaseous mixture when $0.5 L$ of $H_{2}$ at 0.8 bar and $2 L$ of oxygen at 0.7 bar are introduced in $1 L$ vessel at $27^{\circ} C$ ?

1

1.8 bar

2

0.4 bar

3

1.4 bar

4

2.5 bar

Explanation:

Ideal gas equation is $PV = nRT$
Applying ideal gas equation for $H_{2}$ gas,
$0.8 \times 0.5 = n_{H_{2}} R T \Rightarrow n_{H_{2}} = \frac{0.8 \times 0.5}{R T} = \frac{0.4}{R T}$
Applying ideal gas equation for $O_{2}$ gas,
$0.7 \times 2.0 = n_{O_{2}} R T \Rightarrow n_{O_{2}} = \frac{0.7 \times 2.0}{R T} = \frac{1.4}{R T}$
When gas mixture is introduced in $1 L$ vessel, then
$P V = (n_{H_{2}} + n_{O_{2}}) R T$
$P \times 1 = (\frac{0 .4}{R T} + \frac{1 .4}{R T}) R T \Rightarrow P = 0 .4 + 1 .4 = 1 .8 b a r$

CHXI06:STATES OF MATTER

314214 0.5 moles of gas $A$ and $x$ moles of gas $B$ exert a pressure of $200 Pa$ in a container of volume 10 $m^{3}$ at $1000 K$ . Given $R$ is the gas constant in $J K^{- 1} m o l^{- 1} .$ The value of 'x' is

1

\frac{2 R}{4 + R}

2

\frac{2 R}{4 - R}

3

\frac{4 - R}{2 R}

4

\frac{4 + R}{2 R}

Explanation:

Total number of moles $(n_{T}) = (0 .5 + x)$
$PV = nRT$
$200 \times 10 = (0 .5 + x) R \times 1000 \frac{2}{R}$ $= 0 .5 + x$
$\Rightarrow \frac{2}{R} = \frac{1}{2} + x$
$\Rightarrow \frac{2}{R} - \frac{1}{2} = x \Rightarrow x = \frac{4 - R}{2 R}$

JEE - 2018

CHXI06:STATES OF MATTER

314211 The volume of $0.0168 mol$ of $O_{2}$ obtained by decomposition of ${KClO}_{3}$ and collected by displacement of water is $428 mL$ at pressure $754 mm$ of $Hg$ at $25^{\circ} C$ . The pressure of water vapour (in $mm$ of $Hg$ ) at $25^{\circ} C$ is

1 18.5

2 20.6

3 22.3

4 24.6

Explanation:

Volume of $0.0168 mol$ of $O_{2}$ at $STP$
$= 0.0168 \times 22400 mL = 376.3 mL$
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} \Rightarrow \frac{760 \times 376.3}{273} = \frac{P_{2} \times 428}{298}$
$\Rightarrow P_{2} = 729.4 mm$ of $Hg$
Pressure of water vapor $= 754 - 729.4 = 24.6$ $mm$ of $Hg$

CHXI06:STATES OF MATTER

314212 If $1 0^{- 4} d m^{3}$ of water is introduced into a $1 .0 d m^{3}$ flask at $300 K$ , then the number of moles of water formed in the vapour phase when equilibrium is established is [Given: Vapour pressure of $H_{2} O$ at $300 K$ is $3170 Pa$ ; $R = 8.314 {JK}^{- 1} {mol}^{- 1}]$

1

1.27 \times 10^{- 3}

2

5.56 \times 10^{- 3}

3

1.53 \times 10^{- 2}

4

4.46 \times 10^{- 2}

Explanation:

$P V = n R T; P = 3170 Pa; V = 1 {dm}^{3} = 10^{- 3} m^{3}$
$3170 \times 10^{- 3} \times 1 = n \times 8.314 \times 300$
$n = \frac{3170 \times 10^{- 3}}{8.314 \times 300} = 1.27 \times 10^{- 3}$

CHXI06:STATES OF MATTER

314213 What will be the pressure of the gaseous mixture when $0.5 L$ of $H_{2}$ at 0.8 bar and $2 L$ of oxygen at 0.7 bar are introduced in $1 L$ vessel at $27^{\circ} C$ ?

1

1.8 bar

2

0.4 bar

3

1.4 bar

4

2.5 bar

Explanation:

Ideal gas equation is $PV = nRT$
Applying ideal gas equation for $H_{2}$ gas,
$0.8 \times 0.5 = n_{H_{2}} R T \Rightarrow n_{H_{2}} = \frac{0.8 \times 0.5}{R T} = \frac{0.4}{R T}$
Applying ideal gas equation for $O_{2}$ gas,
$0.7 \times 2.0 = n_{O_{2}} R T \Rightarrow n_{O_{2}} = \frac{0.7 \times 2.0}{R T} = \frac{1.4}{R T}$
When gas mixture is introduced in $1 L$ vessel, then
$P V = (n_{H_{2}} + n_{O_{2}}) R T$
$P \times 1 = (\frac{0 .4}{R T} + \frac{1 .4}{R T}) R T \Rightarrow P = 0 .4 + 1 .4 = 1 .8 b a r$

CHXI06:STATES OF MATTER

314214 0.5 moles of gas $A$ and $x$ moles of gas $B$ exert a pressure of $200 Pa$ in a container of volume 10 $m^{3}$ at $1000 K$ . Given $R$ is the gas constant in $J K^{- 1} m o l^{- 1} .$ The value of 'x' is

1

\frac{2 R}{4 + R}

2

\frac{2 R}{4 - R}

3

\frac{4 - R}{2 R}

4

\frac{4 + R}{2 R}

Explanation:

Total number of moles $(n_{T}) = (0 .5 + x)$
$PV = nRT$
$200 \times 10 = (0 .5 + x) R \times 1000 \frac{2}{R}$ $= 0 .5 + x$
$\Rightarrow \frac{2}{R} = \frac{1}{2} + x$
$\Rightarrow \frac{2}{R} - \frac{1}{2} = x \Rightarrow x = \frac{4 - R}{2 R}$

JEE - 2018

CHXI06:STATES OF MATTER

314211 The volume of $0.0168 mol$ of $O_{2}$ obtained by decomposition of ${KClO}_{3}$ and collected by displacement of water is $428 mL$ at pressure $754 mm$ of $Hg$ at $25^{\circ} C$ . The pressure of water vapour (in $mm$ of $Hg$ ) at $25^{\circ} C$ is

1 18.5

2 20.6

3 22.3

4 24.6

Explanation:

Volume of $0.0168 mol$ of $O_{2}$ at $STP$
$= 0.0168 \times 22400 mL = 376.3 mL$
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} \Rightarrow \frac{760 \times 376.3}{273} = \frac{P_{2} \times 428}{298}$
$\Rightarrow P_{2} = 729.4 mm$ of $Hg$
Pressure of water vapor $= 754 - 729.4 = 24.6$ $mm$ of $Hg$

CHXI06:STATES OF MATTER

314212 If $1 0^{- 4} d m^{3}$ of water is introduced into a $1 .0 d m^{3}$ flask at $300 K$ , then the number of moles of water formed in the vapour phase when equilibrium is established is [Given: Vapour pressure of $H_{2} O$ at $300 K$ is $3170 Pa$ ; $R = 8.314 {JK}^{- 1} {mol}^{- 1}]$

1

1.27 \times 10^{- 3}

2

5.56 \times 10^{- 3}

3

1.53 \times 10^{- 2}

4

4.46 \times 10^{- 2}

Explanation:

$P V = n R T; P = 3170 Pa; V = 1 {dm}^{3} = 10^{- 3} m^{3}$
$3170 \times 10^{- 3} \times 1 = n \times 8.314 \times 300$
$n = \frac{3170 \times 10^{- 3}}{8.314 \times 300} = 1.27 \times 10^{- 3}$

CHXI06:STATES OF MATTER

314213 What will be the pressure of the gaseous mixture when $0.5 L$ of $H_{2}$ at 0.8 bar and $2 L$ of oxygen at 0.7 bar are introduced in $1 L$ vessel at $27^{\circ} C$ ?

1

1.8 bar

2

0.4 bar

3

1.4 bar

4

2.5 bar

Explanation:

Ideal gas equation is $PV = nRT$
Applying ideal gas equation for $H_{2}$ gas,
$0.8 \times 0.5 = n_{H_{2}} R T \Rightarrow n_{H_{2}} = \frac{0.8 \times 0.5}{R T} = \frac{0.4}{R T}$
Applying ideal gas equation for $O_{2}$ gas,
$0.7 \times 2.0 = n_{O_{2}} R T \Rightarrow n_{O_{2}} = \frac{0.7 \times 2.0}{R T} = \frac{1.4}{R T}$
When gas mixture is introduced in $1 L$ vessel, then
$P V = (n_{H_{2}} + n_{O_{2}}) R T$
$P \times 1 = (\frac{0 .4}{R T} + \frac{1 .4}{R T}) R T \Rightarrow P = 0 .4 + 1 .4 = 1 .8 b a r$

CHXI06:STATES OF MATTER

314214 0.5 moles of gas $A$ and $x$ moles of gas $B$ exert a pressure of $200 Pa$ in a container of volume 10 $m^{3}$ at $1000 K$ . Given $R$ is the gas constant in $J K^{- 1} m o l^{- 1} .$ The value of 'x' is

1

\frac{2 R}{4 + R}

2

\frac{2 R}{4 - R}

3

\frac{4 - R}{2 R}

4

\frac{4 + R}{2 R}

Explanation:

Total number of moles $(n_{T}) = (0 .5 + x)$
$PV = nRT$
$200 \times 10 = (0 .5 + x) R \times 1000 \frac{2}{R}$ $= 0 .5 + x$
$\Rightarrow \frac{2}{R} = \frac{1}{2} + x$
$\Rightarrow \frac{2}{R} - \frac{1}{2} = x \Rightarrow x = \frac{4 - R}{2 R}$

JEE - 2018