States of Matter Test

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States of Matter Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

A vessel contains equal masses of three gases $$A,\ B,\ C$$ and recorder a pressure of 3.5 bar at $${ 25 }^{ o }$$ C. The molecular mass of $$C$$ is twice that of $$B$$ and molecular mass of $$A$$ is half of B. Find the partial pressure of $$B$$ (in bar the vessel).

A
3

B
2

C
4

D
1

Solution

$$ m_{A} = m_{B} = m_{C} = m \hspace{20mm} P_{total}=3.5 bar \hspace{20mm} M_{C} = 2M_{B} \hspace{20mm} 2M_{A}=M_{B} \\ \text{Partial Pressure of B = } \cfrac{\cfrac{m_{B}}{M_{B}}}{\cfrac{m_{A}}{M_{A}}+\cfrac{m_{B}}{M_{B}} +\cfrac{m_{C}}{M_{C}}} \cdot P_{total} = \cfrac{\cfrac{m}{M_{B}}}{\cfrac{m}{M_{B}/2}+\cfrac{m}{M_{B}} +\cfrac{m}{2M_{B}}} \cdot P_{total} = \cfrac{1}{2+1+\cfrac{1}{2}} \cdot P_{total} \\ \hspace{35mm} = \cfrac{2}{7} \cdot P_{total} = \cfrac{2}{7} \times 3.5 = 1bar $$
Question 2
1 / -0

Two liquids A and B form an ideal solution. The vapour pressure of pure A and pure B are 66mm of Hg and 88mm of Hg, respectively. Calculate the composition of vapour A in the solution which is equilibrium and whose molar volume is 36%.

A
0.43

B
0.70

C
0.30

D
0.50

Solution

$$P_t= P_A+P_B =66mm+88mm=154mm$$ of $$Hg$$
Now, $$P_A$$ $$=X_A \times P_T$$
Therefore, $$X_A$$ =$$\cfrac{P_A}{P_T}=\cfrac{66}{154}$$ $$=0.43$$
Therefore, composition of Vapour A is $$0.43$$.
Question 3
1 / -0

The vapour pressure of pure liquid is $$70$$ Torr at $$27^o C$$. The vapour pressure of a solution of this liquid and another liquid (mole fraction 0.2) is $$84$$ Torr at $$27^o C$$. The vapour pressure of pure liquid $$B$$ at $$27^o C$$ is?

A
$$140$$ Torr

B
$$280$$ Torr

C
$$160$$ Torr

D
$$200$$ Torr

Solution

Vapour Pressure of pure liquid $$A=70$$ torr ($$P_{A}^{o}$$)

Mole fraction of liquid $$B=0.2$$ ($$x_{B}$$)

Thus mole fraction of liquid $$A=x_{A}=1-x_{B}$$

Total vapour pressure $$=84$$ torr

$$P=x_{A} P_{A}^{o} + x_{B} P_{B}^{o}$$

$$84=(0.8) \times 70 + 0.2 \times P_{B}^{o}$$

$$P_{B}^{o}$$ $$=140$$ torr $$=$$ vapour pressure of the pure liquid $$B$$

Hence, the correct option is $$\text{A}$$
Question 4
1 / -0

A solution has a $$1:4$$ mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at $$20^{\circ}C$$ are $$440$$ mm of $$Hg$$ for pentane and $$120$$ mm of $$Hg$$ for hexane. The mole fraction of pentane in the vapour phase would be?

A
$$0.549$$

B
$$0.200$$

C
$$0.786$$

D
$$0.478$$

Solution

Mole fraction of pentane solution= $$\cfrac {1}{1+4}=\cfrac {1}{5}$$

Mole fraction of hexane solution= $$\cfrac {4}{1+4}=\cfrac {4}{5}$$

Total pressure of solution, $$P_S=X_{P}P_P^O+X_HP_H^O$$

$$=\cfrac {1}{5}\times 440\ mm\ Hg+\cfrac {4}{5}\times 120\ mm\ Hg$$
$$=184\ mm\ Hg$$

$$\therefore$$ The fraction of pentane in the vapour phase

$$=\cfrac {X_PP_P^O}{P_S}=\cfrac {88}{184}=0.478$$ .

Hence, the correct option is $$D$$
Question 5
1 / -0

$${ \Delta }_{ f }{ G }^{ o }$$ at $$500\ K$$ for substance $$S$$ in liquid state and gaseous state are $$ +100.7\ kcal\ { mol }^{ -1 }$$ and $$ +103\ kcal\ { mol }^{ -1 }$$ respectively. Vapour pressure of liquid $$S$$ at $$500\ K$$ is approximately equal to: $$(R= 2\ { cal }\ K^{ -1 }\ { mol }^{ -1 })$$

A
$$0.1\ atm$$

B
$$1\ atm$$

C
$$10\ atm$$

D
$$100\ atm$$

Solution

Substance (s) $$\rightarrow$$ Substance (g)
$$\triangle G^{o}=103-100.7$$
$$=2.3\ K\ cal\ mol^{-1}$$

$$K_{p}=$$ Vapour pressure of the substance
$$K_{p}=P$$

$$\triangle G=-RT\ In\ K_{p}$$ $$(\because K_{p}=P)$$

$$2.3=-2\times 10^{-3}\times 500\ In\ P$$

$$2.3=-2\times 500\times10^{-3}\times 2.3\ \log P$$

$$\log P=-1$$

$$P=0. 1\ atm$$
Question 6
1 / -0

The vapour pressure of a solvent decreased by 10 mm of mercury when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be 20 mm of mercury?

A
0.4

B
0.6

C
0.8

D
0.2

Solution

$$\cfrac {\Delta P}{P_o}=\cfrac {n}{n+N}$$

$$\Rightarrow \cfrac {10}{P_o}=0.2 $$

$$\Rightarrow P_o=50mm$$

For other solutions of same solvent,

$$\Rightarrow \cfrac {20}{P_o}=\cfrac {n}{n+N}$$

$$\Rightarrow \cfrac {20}{50}=$$ mole fraction of solute= $$0.4$$

$$\therefore$$ Mole fraction of solvent= $$1-0.4=0.6$$

Hence, the correct option is $$\text{B}$$
Question 7
1 / -0

$$20$$ litre of an ideal gas, present in a piston fitted cylinder at $$10$$ atm is allowed to expand in a process in which $$P/V$$ is a constant. If final volume of gas is $$60$$ litre, then work done by gas is:

A
$$81.12\ kJ$$

B
$$-81.12\ kJ$$

C
$$40.56\ kJ$$

D
$$-40.56\ kJ$$

Solution

$$W=-P(V_2-V_1)$$
$$=10(60-20)$$
$$=10(40)=-400 atmL$$
$$1 atm= 101325 Pa$$
$$\Rightarrow 1 atm= 101325J/m^3$$
$$\Rightarrow 1L= 10^{-3}m^3$$
$$\therefore W= 400 \times 101325 \cfrac {J}{M^3}\times 10^{-3}M^{3}$$
$$=-40.56 kJ$$
Question 8
1 / -0

A bulb of unknown volume contained an ideal gas at 650mm pressure. A certain amount of the gas was withdrawn and found to occupy 1.5 cc at 700 mm pressure. The pressure of the gas remaining in the bulb was found to be 600 mm. If all measurements are made at the same temperature, the volume of the bulb is ?

A
21 cc

B
42 cc

C
105 cc

D
63 cc

Solution

The number of moles of gas withdrawn, $$n_1=\cfrac {RT}{PV}=\cfrac {RT}{1.5 cc\times 700 mm}=\cfrac {RT}{1050}$$
If the total volume of bulb is $$x$$ cc
The number of moles of gas remaining in the bulb, $$n_2=\cfrac {RT}{x\times 600}$$
Again, before withdrawal of gas
$$n_1+n_2=\cfrac {RT}{x\times 650}$$
$$\Rightarrow RT\left(\cfrac {1}{1050}+\cfrac {1}{600 \times x}\right)=\cfrac {RT}{x\times 650}$$
$$\Rightarrow \cfrac {1}{1050}=\cfrac {1}{x}\left(\cfrac {1}{650}-\cfrac {1}{600}\right) \Rightarrow x= 21$$ cc
Question 9
1 / -0

A container is filled with gas to a pressure of 2 atm at $$0^o$$ C. At what temperature will the pressure inside the container be 2.50 atm?

A
$$75^o C$$

B
$$12.5^o C$$

C
$$18^o C$$

D
$$68.25^o C$$

Solution

Answer : D. 68.25$$^{o}C$$

We have ideal gas equation, $$PV = nRT$$

Here, only P and T changes. n, R and V (volume of the container) remain same.

From the question,
$$P_{1}$$ = 2 atm
$$T_{1}$$ = $$0^{o}\textrm{C}$$ = $$273K$$

$$P_{2}$$ = 2.50 atm
$$T_{2}$$ = ?

For $$P_{1}$$ and $$T_{1}$$ ideal gas equation becomes, $$P_{1}V = nRT_{1}$$ ---------> ( 1 )

For $$P_{2}$$ and $$T_{2}$$ ideal gas equation becomes, $$P_{2}V = nRT_{2}$$ ---------> ( 2 )

Divide equation ( 1 ) by ( 2 ) we get,

$$\dfrac{P_{1}V}{P_{2}V} = \dfrac{nRT_{1}}{nRT_{2}}$$

$$=>\dfrac{P_{1}}{P_{2}} = \dfrac{T_{1}}{T_{2}}$$ $$=>\dfrac{2}{2.50} = \dfrac{273}{T_{2}}$$

$$=>T_{2} = 273\times \dfrac{2.50}{2} = 341.25K$$
$$=>T_{2} = 341.25-273 ^{o}C = 68.25^{o}C$$

option D. $$68.25^{o}C$$ is correct
Question 10
1 / -0

An open bulb containing air at $$19^{\circ}C$$ was cooled to a certain temperature at which the number of moles of the gaseous molecules increased by $$25\%$$. The final temperature is:

A
$$-39.4^{\circ}C$$

B
$$233.6^{\circ}C$$

C
$$39.4^{\circ}C$$

D
$$240^{\circ}C$$

Solution

From the ideal gas relation,

$$PV=nRT$$

Since the pressure and volume in the given problem is constant, we get

$$nT=constant$$

$$\Rightarrow n_1T_1=n_2T_2$$

$$\Rightarrow T_2=\dfrac{T_1}{1.25}=\dfrac{292}{1.25}=233.6\ K=-39.4^o\ C$$

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Re-Attempt Weekly Quiz Competition

States of Matter Test - 38

Result

States of Matter Test - 38

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SHARING IS CARING

Question 1

3

2

4

1

Solution

Question 2

0.43

0.70

0.30

0.50

Solution

Question 3

$$140$$ Torr

$$280$$ Torr

$$160$$ Torr

$$200$$ Torr

Solution

Question 4

$$0.549$$

$$0.200$$

$$0.786$$

$$0.478$$

Solution

Question 5

$$0.1\ atm$$

$$1\ atm$$

$$10\ atm$$

$$100\ atm$$

Solution

Question 6

0.4

0.6

0.8

0.2

Solution

Question 7

$$81.12\ kJ$$

$$-81.12\ kJ$$

$$40.56\ kJ$$

$$-40.56\ kJ$$

Solution

Question 8

21 cc

42 cc

105 cc

63 cc

Solution

Question 9

$$75^o C$$

$$12.5^o C$$

$$18^o C$$

$$68.25^o C$$

Solution

Question 10

$$-39.4^{\circ}C$$

$$233.6^{\circ}C$$

$$39.4^{\circ}C$$

$$240^{\circ}C$$

Solution

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