States of Matter Test

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States of Matter Test - 26

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Question 1
1 / -0

On a humid day in summer, the mole fraction of gaseous $$H_{2}O$$ (water vapour) in the air at $$25^0$$C can be as high as $$0.0287$$. Assuming a total pressure of $$0.977$$ atm. What is the partial pressure of dry air?

A
$$94.9$$ atm

B
$$0.949$$ atm

C
$$949$$ atm

D
$$0.648$$ atm

Solution

$$p_{H_2O}=X_{H_2O} p_{\text{total}}$$

$$ =0.0287\times 0.977 =0.028atm$$

$$p_{\text{total}}=p_{\text{dryair}}+p_{H_2O}$$

$$ p_{\text{dryair}}=p_{\text{total}}−p_{H_2O} =0.977−0.028=0.949 atm$$
Question 2
1 / -0

The ratio of the vapour pressure of a solution to the vapour pressure of the solvent is:

A
equal to the mole fraction of the solvent

B
equal to the mole fraction of the solute

C
directly proportional to mole fraction of the solute

D
None of the above
Question 3
1 / -0

At a given temperature, total vapour pressure in Torr of a mixture volatile components A and B is given by
$$P_{Total}=120-75X_B$$
Hence, vapour pressure of pure A and B respectively (in Torr) are ?

A
120, 75

B
120, 195

C
120, 45

D
75, 45

Solution

$$\displaystyle P_{Total}=P_AX_A+P_BX_B$$
But $$\displaystyle X_A=1-X_B$$
$$\displaystyle ( \because X_A+X_B=1)$$
Hence, $$\displaystyle P_{Total}=P_A(1-X_B)+P_BX_B$$
$$\displaystyle P_{Total}=P_A-P_AX_B+P_BX_B$$
$$\displaystyle P_{Total}=P_A-(P_A-P_B)X_B$$
But, $$\displaystyle P_{Total}=120-75X_B $$
Hence, $$\displaystyle P_A=120 \ torr$$
$$\displaystyle P_A-P_B = 75$$
$$\displaystyle P_B=P_A-75$$
$$\displaystyle P_B=120-75$$
$$\displaystyle P_B=45 \ torr$$
Question 4
1 / -0

Pressure exerted by one mole of an ideal gas kept in vessel of '$$V$$' L having root mean square molecule '$$V$$' and '$$m$$' mass of each molecule is correctly given by the equation?

A
$$P =$$ $$\dfrac {1}{2}$$$$\dfrac {N}{V}$$mv$$^{2}$$

B
$$P =$$ $$\dfrac {1}{3}$$$$\dfrac {N}{V}$$mv$$^{2}$$

C
$$P =$$ $$\dfrac {2}{3}$$$$\dfrac {N}{V}$$mv$$^{2}$$

D
$$P =$$ $$\dfrac {3}{2}$$$$\dfrac {N}{V}$$mv$$^{2}$$

Solution

According to ideal gas and kinetic theory of gases, pressure exerted by one mole of ideal gas in a vessel of volume $$V$$, having mean square molecule $$'V'$$ and $$'m'$$ mass of each molecule is given by
$$P=\cfrac { 1 }{ 3 } \cfrac { N }{ V } { mv }^{ 2 }$$
Question 5
1 / -0

Water and chlorobenzene are immiscible liquids. Their mixture boils at $$89^o$$C under a reduced pressure of $$7.7 \times 10^4$$ Pa. The vapour pressure of pure water at $$89^o$$C is $$7 \times 10^4$$ Pa. Weight percent of chlorobenzene in the distillate is:

A
50

B
60

C
79

D
38.46

Solution

Vapour pressure = $$7.7 \times 10^4$$ Pa
Total pressure = $$7 \times 10^4$$ Pa
Partial pressure = Vapour pressure = Mole fraction x Total pressure
Mole fraction of $$H_2O$$ = $$\dfrac{Vapour pressure}{Total pressure}$$ = $$\dfrac{7.7 \times 10^4}{7 \times 10^4}$$ = 0.909
Mole fraction of $$CHCl_3$$ = 1 - 0.909 = 0.091
Molar mass of $$H_2O$$ = 18g/mol
Molar mass of $$CHCl_3$$ = 119.5g/mol
Weight percent of chloroform = Mole fraction of $$CHCl_3$$ x Molar mass of $$CHCl_3$$ / [Molar mass of $$CHCl_3$$ +Molar mass of $$H_2O$$ }
Weight percent of chloroform = 79%
Answer is 79%.
Question 6
1 / -0

Two glass bulbs A and B are connected by a very small tube having a stop cock. Bulb A has a volume of 100 $$cm^{3}$$ and contained the gas, while bulb B was empty. On opening the stop cock, the pressure fell down 40%. The volume of the bulb B must be ?

A
75 $$cm^{3}$$

B
125 $$cm^{3}$$

C
150 $$cm^{3}$$

D
250 $$cm^{3}$$

Solution

Let the initial pressure in bulb A be P.
Given that the pressure fell down $$40 \%$$ on opening the stop cock
$$\therefore$$ Final pressure $$= \cfrac{40}{100} P = \cfrac{2}{5} P$$
Initial voume of bulb A $$= 100{cm}^{3}$$
Let initial volume of bulb B be V.
$$\therefore$$ total volume after opening the stopcock $$= 100 + V$$
By using Boyle's law,
$${P}_{i}{V}_{i} = {P}_{f}{V}_{f}$$
$$P \times 100 = \cfrac{2}{5} P \times \left( 100 + V \right)$$
$$\Rightarrow \; V = 150 {cm}^{3}$$
Hence, the volume of bulb B must be $$150{cm}^{3}$$.
Question 7
1 / -0

At $$80^0$$C, the vapour pressure of pure liquid A is $$250$$mm of Hg and that of pure liquid B is $$1000$$ mm of Hg. If a solution of A and B boils at $$80^0$$C and $$1$$ atm pressure, the amount of A in the mixture is :$$(1atm = 760mm$$Hg)

A
$$50$$ mole percent

B
$$52$$ mole percent

C
$$32$$ mole percent

D
$$48$$ mole percent

Solution

1 atm = 760 mm Hg = $$P_T$$
$$P_T = P^0_A X_A \times P^0_BX_B$$
$$760 = 250X_A + 1000(1 - X_A)$$
$$ 240 = 750X_A$$
$$X_A = 0.32$$
Hence mole $$\%$$ of A = $$32\%$$
Question 8
1 / -0

Which of the following statement is wrong ?

A
For gas A, a= 0 and z will linearly depend on pressure

B
For gas B, b = 0 and z will linearly depend on pressure

C
Gas C is real gas and we can find "a" and 'b' if intersection data is given

D
All vander waal gases will behave like gas C and give positive slope at high pressure

Solution

According to the real gas equation, and characteristic of constants 'a' and 'b'

$$(P-\cfrac{n^2a}{V^2}) (V-nb)=RT$$

$$Z=\cfrac{PV}{RT}$$

For gas A, a= 0 and z will linearly depend on pressure as per the below relation-
$$Z=\cfrac{P(V-nb)}{RT}$$

For gas B, b= 0 and z will linearly depend on the volume, not pressure as per the below relation-

$$Z=\cfrac{(P-\cfrac{n^2a}{V^2})V}{RT}$$

The dotted plot given by gas C is for real gases.

At higher pressure Z>1.

Hence, statement B is incorrect.
Question 9
1 / -0

At $$80^o$$C, the vapour pressure of pure liquid $$A$$ is $$250$$ mm of $$Hg$$ and that of pure liquid $$B$$ is $$1000$$ mm of $$Hg$$. If a solution of $$A$$ and $$B$$ boils at $$80^o$$ and $$1$$ atm pressure, the amount of $$A$$ in the mixture is?
($$1$$ atm $$=760$$ mm Hg).

A
$$50$$ mole percent

B
$$52$$ mole percent

C
$$32$$ mole percent

D
$$48$$ mole percent

Solution

1 atm = 760 mm Hg = $$P_T$$

$$P_T = P^0_A X_A \times P^0_BX_B$$

$$760 = 250X_A + 1000(1 - X_A)$$

$$ 240 = 750X_A$$

$$X_A = 0.32$$

Hence mole $$\%$$ of A $$= 32\%$$
Question 10
1 / -0

At 80C, the vapour pressure of pure benzene is 753 mm Hg and of pure toluene 290 mm Hg. Calculate the composition of a liquid in mole percent which at $$80$$ is in equilibrium with vapour containing 30 mole percent of benzene.

A
24, 56

B
26 ,54

C
56 24

D
34 ,76

Solution

Mole % of benzene=30%of solution
Moles of benzene=30*80/100=24
Moles of Toluene=70*80/100=56
Composition of benzene and Toluene are 24 moles and 56 moles respectively.

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States of Matter Test - 26

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States of Matter Test - 26

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Question 1

$$94.9$$ atm

$$0.949$$ atm

$$949$$ atm

$$0.648$$ atm

Solution

Question 2

equal to the mole fraction of the solvent

equal to the mole fraction of the solute

directly proportional to mole fraction of the solute

None of the above

Question 3

120, 75

120, 195

120, 45

75, 45

Solution

Question 4

$$P =$$ $$\dfrac {1}{2}$$$$\dfrac {N}{V}$$mv$$^{2}$$

$$P =$$ $$\dfrac {1}{3}$$$$\dfrac {N}{V}$$mv$$^{2}$$

$$P =$$ $$\dfrac {2}{3}$$$$\dfrac {N}{V}$$mv$$^{2}$$

$$P =$$ $$\dfrac {3}{2}$$$$\dfrac {N}{V}$$mv$$^{2}$$

Solution

Question 5

50

60

79

38.46

Solution

Question 6

75 $$cm^{3}$$

125 $$cm^{3}$$

150 $$cm^{3}$$

250 $$cm^{3}$$

Solution

Question 7

$$50$$ mole percent

$$52$$ mole percent

$$32$$ mole percent

$$48$$ mole percent

Solution

Question 8

For gas A, a= 0 and z will linearly depend on pressure

For gas B, b = 0 and z will linearly depend on pressure

Gas C is real gas and we can find "a" and 'b' if intersection data is given

All vander waal gases will behave like gas C and give positive slope at high pressure

Solution

Question 9

$$50$$ mole percent

$$52$$ mole percent

$$32$$ mole percent

$$48$$ mole percent

Solution

Question 10

24, 56

26 ,54

56 24

34 ,76

Solution

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