States of Matter Test

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States of Matter Test - 59

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Question 1
1 / -0

CuSO$$_4.$$5$$H_2O$$(s) $$\rightleftharpoons$$ CuSO$$_4$$.3H$$_2O$$(s) + 2H$$_2O$$(g); $$k_p$$ = 4 $$10^{4}$$ atm$$^2.$$ If the vapour pressure of water is 38 torr then percentage of relative humidity is : (Assume all data at constant temperature)

A
4

B
10

C
40

D
None of these

Solution

$$ \begin{aligned} K_ P &=\left(P_{ H_{2} O}\right)^{2} \\ P_{ H_{2} O }&=\sqrt{K_ P} \\ &=\sqrt{4 \times 10^{-4}} \\ &=2 \times 10^{-2} \end{aligned} $$

$$ \text { % Relative Humidity }=\dfrac{P_{H_2O} \times 760}{\text { v.p of water}} $$

$$ \begin{aligned} =& \dfrac{2 \times 10^{-2} \times 760 \times 100}{38} \\ &=40 \end{aligned} $$
Question 2
1 / -0

Vapour Pressure of a mixture of benzene and toluene is given by $$P= 179X_{B} + 92$$, Where $$X_{B}$$ is mole fraction of benzene.
This condensed liquid again brought to the same temperature then what will be the mole fraction of benzene in vapour phase :

A
0.007

B
0.93

C
0.65

D
4.5

Solution

Mole fraction of $${ C }_{ 6 }{ H }_{ 6 }$$ in vapour phase $$={ X }_{ B }^{ 1 }$$ of initial mixture
$${ X }_{ B }^{ 1 }=\dfrac { { P }_{ B }^{ 1 } }{ { P }_{ M } } =\dfrac { 162.6 }{ 199.4 } =0.815$$
then mole fraction of $${ C }_{ 7 }{ H }_{ 8 }=0.185$$
(mole fraction of $${ C }_{ 6 }{ H }_{ 6 }$$ in initial mixture) $$=$$ (mole fraction of $${ C }_{ 6 }{ H }_{ 6 }$$ in liquid phase on $$II$$ mixture $${ X }_{ B }^{ 1 }$$)
$${ P }_{ M }={ P }_{ B }^{ 1 }+{ P }_{ T }^{ 1 }$$
now, $${ P }_{ M }={ P }_{ B }^{ 0 }{ X }_{ B }^{ 1 }+{ P }_{ T }^{ 0 }{ X }_{ T }^{ 1 }$$
$$=\left[ \left( 271 \right) \left( 0.815 \right) +\left( 92 \right) \left( 0.185 \right) \right] $$ mm
$$=237.885$$ mm
new mole fraction of $${ C }_{ 6 }{ H }_{ 6 }=\dfrac { 220.865 }{ 237.885 } =0.928\cong 0.93$$
$$\therefore$$ new mole fraction of Benzene in vapour phase is $$0.93$$.
Question 3
1 / -0

What is the effect on the pressure of a gas if its temperature is increased at constant volume?

A
The pressure of the gas increases

B
The pressure of the gas decreases

C
The pressure of the gas remains same

D
The pressure of the gas becomes double

Solution

According to Gay Lussac's law, at constant volume, the pressure of the gas is directly proportional to the temperature.
$$\therefore P\propto T$$
Thus, with an increase in temperature at constant volume the pressure of the gas increases.
Question 4
1 / -0

$$100$$ grams of oxygen $$(O_{2})$$ gas and $$100$$ grams of helium $$(He)$$ gas are in separate containers of equal volume at $$100^{\circ}C$$. Which one of the following statement is correct?

A
Both gases would have the same pressure

B
The average kinetic energy of $$O_{2}$$ molecules is greater than that of $$He$$ molecules

C
The pressure of $$He$$ gas would be greater than that of the $$O_{2}$$ gas

D
The average kinetic energy of $$He$$ and $$O_{2}$$ molecules is same

Solution

According to ideal gas equation,
$$PV = nRT$$
For $$100\ gm\ O_{2}$$,
$$P_{1}\times V = \dfrac {100}{32} \times R\times T ..... (1)$$
For $$100\ gm\ He$$,
$$P_{2}\times V = \dfrac {100}{4}\times R\times T .... (2)$$
Dividing eq $$(1)$$ by $$(2)$$
$$\dfrac {P_{1}\times V}{P_{2}\times V} = \dfrac {100\times R\times T}{32}\times \dfrac {4}{100}\times \dfrac {1}{R\times T}$$
$$\dfrac {P_{1}}{P_{2}} = \dfrac {1}{8}, P_{2} = 8P_{1}$$
$$\therefore P_{He} > P_{O_{2}}$$.
Question 5
1 / -0

What is a weight of $$CO_2$$ in a $$10L$$ cylinder at $$5atm$$ and $$27^0C$$?

A
$$200\ g$$

B
$$224\ g$$

C
$$44\ g$$

D
$$89.3\ g$$

Solution

$$PV = nRT = \dfrac {m}{M}RT$$
$$P = 5\ atm, V = 10\ L, T = 300\ K$$,
$$R = 0.0821\ L\ atm\ K^{-1} mol^{-1}$$
$$m = \dfrac {PVM}{RT} = \dfrac {5\times 10\times 44}{0.0821\times 300} = 89.3\ g$$.
Question 6
1 / -0

$$X,Y$$ and $$Z$$ in the given graph are?

A
$$X={p}_{1}+{p}_{2},Y=1,Z=0$$

B
$$X={p}_{1}+{p}_{2},Y=0,Z=0$$

C
$$X={p}_{1}\times {p}_{2},Y=0,Z=0$$

D
$$X={p}_{1}-{p}_{2},Y=1,Z=0$$

Solution

Here, $$X =P_T=p_1+p_2=$$ Total pressure of the solution.

$$p_1$$ and $$p_2$$ is the partial pressure of component 1 and 2 respectively.

$$Y=x_2=$$ mole fraction of component 2 in the liquid phase.

$$Z=y_2=$$ mole fraction of component 2 in the vapour phase.

When mole fraction of component 2 in the liquid phase is 0 then mole fraction of component 2 in the vapour phase is also 0 which means the liquid is pure of component 1 only.

So, the correct option is B.
Question 7
1 / -0

A container of $$1\ L$$ capacity contains a mixture of $$4\ g$$ of $$O_{2}$$ and $$2\ g$$ of $$H_{2}$$ at $$0^{\circ}C$$. What will be the total pressure of the mixture?

A
$$50.42\ atm$$

B
$$25.21\ atm$$

C
$$15.2\ atm$$

D
$$12.5\ atm$$

Solution

Applying $$PV = \dfrac {w}{M}RT$$
$$p_{O_{2}} = \dfrac {4}{32}\times \dfrac {0.0821\times 273}{1} = 2.81\ atm$$
$$p_{H_{2}} = \dfrac {2}{2}\times \dfrac {0.0821\times 273}{1} = 22.4\ atm$$
$$p_{total} = p_{O_{2}} + p_{H_{2}} = 2.81 + 22.4 = 25.21\ atm$$.
Question 8
1 / -0

A plot of volume $$(V)$$ versus temperature $$(T)$$ for a gas at constant pressure is a straight line passing through the origin. The plots at different values of pressure are shown in figure. Which of the following order of pressure is correct for this gas?

A
$$P_1>P_2>P_3>P_4$$

B
$$P_1=P_2=P_3=P_4$$

C
$$P_1 < P_2 < P_3 < P_4$$

D
$$P_1 < P_2 = P_3 < P_4$$

Solution

$$\bf{Explanation-}$$
From ideal gas equation:
$$PV = nRT$$
Where $$n$$ is the number of moles.
$$R$$ is redberg's constant.
$$P$$ is pressure
$$V$$ is Volume
$$T$$ is Absolute Temperature
Here, $$n$$ and $$R$$ is constant.
So, $$PV = T$$
Or $$P = \dfrac{T}{V}$$
That means $$P$$ is directly proportional to $$T$$ and inversely proportional to $$V$$.
According to graph:
$$V_1>V_2>V_3>V_4$$
Hence, $$P_1<P_2<P_3<P_4$$ $$\left(\because P=\dfrac{1}{V}\right)$$

$$\bf{Conclusion-}$$ Hence, option C is correct.
Question 9
1 / -0

If the ratio of masses of $$SO_{3}$$ and $$O_{2}$$ gases confined in a vessel is $$1 : 1$$, then the ratio of their partial pressures would be :

A
$$5 : 2$$

B
$$2 : 5$$

C
$$2 : 1$$

D
$$1 : 2$$

Solution

Let $$m$$ be the mass of $$SO_{3}$$ and $$O_{2}$$ enclosed in the vessel.
Number of moles of $$SO_{3} = \dfrac {m}{80}$$

Number of moles of $$O_{2} = \dfrac {m}{32}$$

Partial pressure of $$SO_{3}, P_{A} = \dfrac {m}{80}\times \dfrac {R\times T}{V}$$

Partial pressure of $$O_{2}, P_{B} = \dfrac {m}{32}\times \dfrac {R\times T}{V}$$

Now, $$\dfrac {P_{A}}{P_{B}} = \dfrac {m}{80}\times \dfrac {32}{m} = \dfrac {2}{5}$$

Hence, the ratio of partial pressure of $$SO_{3}$$ and $$O_{2}$$ is $$2 : 5$$.
Question 10
1 / -0

A $$2\ L$$ vessel is filled with air at $$50^{\circ}C$$ and a pressure of $$3\ atm$$. The temperature is now raised to $$200^{\circ}C$$. A valve is now opened so that the pressure inside drops to one atm. What will be the fraction of the total number of moles, inside escaped on opening the valve? (Assume no change in the volume of the container).

A
$$7.7$$

B
$$9.9$$

C
$$8.9$$

D
$$0.77$$

Solution

Given that, $$V = 2\ L, T_{1} = 50 + 273 = 323\ K, P_{1} = 3\ atm$$
On heating $$V = 2L, T_{2} = 200 + 273 = 473, P_{2} = ?$$
Using $$P-T\ law, \dfrac {P_{1}}{T_{1}} = \dfrac {P_{2}}{T_{2}}$$
$$P_{2} = \dfrac {3\times 473}{323} = 4.39\ atm$$
and $$n = \dfrac {PV}{RT} = \dfrac {3\times 2}{0.0821\times 323} = 0.226$$
Now valve is opened till the pressure is maintained at $$1\ atm$$.
Thus, at constant $$V$$ and $$T, P\propto n$$
$$\therefore 4.39\propto 0.226$$
$$\therefore 1\propto n_{left}$$
$$\therefore n = \dfrac {0.226}{4.39} = 0.052$$
$$\therefore$$ Moles escaped out $$= 0.226 - 0.052 = 0.174$$
$$\therefore$$ Fraction of moles escaped out $$= \dfrac {0.174}{0.226} = 0.77$$

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Re-Attempt Weekly Quiz Competition

States of Matter Test - 59

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States of Matter Test - 59

Score

-

Rank

-

SHARING IS CARING

Question 1

4

10

40

None of these

Solution

Question 2

0.007

0.93

0.65

4.5

Solution

Question 3

The pressure of the gas increases

The pressure of the gas decreases

The pressure of the gas remains same

The pressure of the gas becomes double

Solution

Question 4

Both gases would have the same pressure

The average kinetic energy of $$O_{2}$$ molecules is greater than that of $$He$$ molecules

The pressure of $$He$$ gas would be greater than that of the $$O_{2}$$ gas

The average kinetic energy of $$He$$ and $$O_{2}$$ molecules is same

Solution

Question 5

$$200\ g$$

$$224\ g$$

$$44\ g$$

$$89.3\ g$$

Solution

Question 6

$$X={p}_{1}+{p}_{2},Y=1,Z=0$$

$$X={p}_{1}+{p}_{2},Y=0,Z=0$$

$$X={p}_{1}\times {p}_{2},Y=0,Z=0$$

$$X={p}_{1}-{p}_{2},Y=1,Z=0$$

Solution

Question 7

$$50.42\ atm$$

$$25.21\ atm$$

$$15.2\ atm$$

$$12.5\ atm$$

Solution

Question 8

$$P_1>P_2>P_3>P_4$$

$$P_1=P_2=P_3=P_4$$

$$P_1 < P_2 < P_3 < P_4$$

$$P_1 < P_2 = P_3 < P_4$$

Solution

Question 9

$$5 : 2$$

$$2 : 5$$

$$2 : 1$$

$$1 : 2$$

Solution

Question 10

$$7.7$$

$$9.9$$

$$8.9$$

$$0.77$$

Solution

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