States of Matter Test

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States of Matter Test - 57

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Question 1
1 / -0

$$P_A \, = \, X_AP_A$$ and $$P_B \, = \, X_BP_B$$
$$P_T \, = \, X_AP_A \, + \, X_BP_B$$
Vapour pressure of mixtures of Benzene $$(C_6H_6)$$ and toluene $$(C_7H_8)$$ at $$50^{\circ}C$$ are given by $$P_M \, = \, 179X_B \, + \, 92$$ where $$X_B$$ is mole fraction of $$C_6H_6$$.
What is the vapour pressure of pure liquids?

A
$$P_B \, = \, 92 mm, \, P_T \, = \, 179 mm$$

B
$$P_B \, = \, 271 mm, \, P_T \, = \, 92 mm$$

C
$$P_B \, = \, 180 mm, \, P_T \, = \, 91 mm$$

D
None of these

Solution

as $$x_A+x_B=1$$
$$PT=x_AP_A+x_BP_B= (1-xB)P_A+x_BP_B$$
$$P_T=P_A+x_B(P_B-P_A) \longrightarrow (1)$$
Given, $$P_M= 92+179x_B\longrightarrow (2)$$
on comparing equation $$(1)$$ & $$(2)$$ we have
$$P_A= 92 mm, P_B-P_A= 179mm$$
$$P_B= 179+P_A=179+92$$
$$P_A=92 mm, P_B=271mm$$
Question 2
1 / -0

If $${ PCl }_{ 5 }$$ is 80% dissociated at $$250$$ then its vapour density at room temperature will be:

A
56.5

B
104.25

C
101.2

D
52.7
Question 3
1 / -0

The vapour pressure of two pure liquids $$A$$ and $$B$$, that form an ideal solution are $$100$$ and $$900$$ torr respectively at temperature $$T$$. This liquid solution of $$A$$ and $$B$$ is composed of $$1\ mole$$ of $$A$$ and $$1\ mole$$ of $$B$$. What will be the pressure, when $$1$$ mole of mixture has been vapourized?

A
$$800\ torr$$

B
$$500\ torr$$

C
$$300\ torr$$

D
None of these

Solution

$$P_A^o = 100 \, torr \,\,\,\, P_B^o = 900 \, torr$$
$$n_A = 1 \, mole \,\,\,\, n_B = 1 \, mol$$
$$x_A = \dfrac{1}{1 + 1} = 0.5 mole$$
$$x_B = \dfrac{1}{1 + 1} = 0.5 mole$$
Pressure $$= P_A^o x_A + P_B^o x_B$$
$$= 100 \times 0.5 + 900 \times 0.5 = 50 +450 = 500 \, torr$$
Question 4
1 / -0

Two containers, X and Y at 300K and 350K with water vapour pressures 22 mm and 40 mm respectively a are connected, initially closed with a valve. If the valves opened.

A
The final pressure in each container is 31 mm

B
The final pressure in each container is 40 mm

C
Mass of $$H_{2}O$$(l) in X increases

D
Mass of $$H_{2}O$$(l) in Y decreases roar.

Solution

$$ \begin{array}{l} \text { - The vapour pressur in the cointainer } y \text { is higher } \\ \text { than the vapour pressure in the container } x \text { . } \\ \text { - When the value is opened, the vapour from } \\ \text { container y diffuse in the container } x \text { } \\ \text { and condense there. } \\ \text { - When the equilibrium is re-established, the } \\ \text { mass of liquid water in the contaier y } \\ \text { decreases and container } x \text { increases. } \\ \text { Hence the answer is, option ( } A \text { ) } \\ \text { i.e. The final pressure in each container is } 31 \mathrm{~mm} \text { . } \end{array} $$
Question 5
1 / -0

The vapour pressure of two pure liquids A and B which form an ideal solution are 1000 and 1600 torr respectively at 400 K. A liquid solution of A and B for which the mole fraction of A is 0.60 is contained in a cylinder by a piston on which the pressure can be varied. The solution use slowly vapourised at 400 K by decreasing the applied pressure. What is the composition of last droplet of liquid remaining in equilibrium with vapour?

A
$$X_A= 0.30, X_B= 0.70$$

B
$$X_A= 0.40, X_B= 0.60$$

C
$$X_A= 0.70, X_B= 0.30$$

D
$$X_A= 0.50, X_B= 0.50$$

Solution

$$\begin{array}{l}\text{Here,} Y_A =\dfrac{ P_A^0 X_A }{P_A^0 X_A+ P^0_B(1-X_A)}\implies 0.6 =\dfrac{1000 X_A}{1000X_A+1600(1-X_A)}\\ \quad \therefore X_A=0.70 \quad X_B = 1-X_A =0.30 \end{array}$$
Question 6
1 / -0

Which of the following is a correct graph?

A

B

C

D

Solution

According to kinetic theory of gas kinetic energy only depending on temperature $$K=\dfrac{3}{2}K_B T$$
Hence option A is correct
Question 7
1 / -0

Two glass bulbs $$A$$ (of $$100\ mL$$ capacity), and $$B$$ (of $$150\ mL$$ capacity) containing same gas are connected by a small tube of negligible volume. At particular temperature, the pressure in $$A$$ was found to be $$20$$ times more than that in bulb $$B$$. The stopcock is opened without changing the temperature. The pressure in $$A$$ will :

A
Drop by $$75$$%

B
Drop $$57$$%

C
Drop by $$25$$%

D
Will remain same

Solution

Let the original pressure in B be P.
$$\therefore$$ Pressure in A = 20 P.

Given that :

Volume in A = 100 mL
Volume in B = 150 mL

After opening the stopcock,
total volume $$= \left( 100 + 150 \right) mL = 250 mL$$

According to Boyle's law :

$${P}_{f}{V}_{f} = {P}_{i}{V}_{i}$$

Let the partial pressure of gas in A be $${P}_{A}$$ :

$${P}_{A} \times 250 = 100 \times 20 P$$

$$\Rightarrow \; {P}_{A} = \cfrac{100 \times 20P}{250} = 8 P$$

Let Partial pressure of gas in B be $${P}_{B}$$ :

$${P}_{B} \times 250 = 150 \times P$$

$${P}_{B} = \cfrac{150 P}{250} = 0.6 P$$

$$\therefore$$ Total pressure $$= 8 P + 0.6 P = 8.6 P$$

Now,

Drop in pressure in A $$= 20P - 8.6P = 11.4 P$$

$$\therefore \; \%$$ drop in pressure $$= \cfrac{11.4 P}{20 P} \times 100 = 57 \%$$

Therefore, the correct option is B.
Question 8
1 / -0

Two liquids $$X$$ and $$Y$$ are perfectly immiscible. If $$X$$ and $$Y$$ have molecular masses in ration $$1 : 2$$, the total vapour pressure of a mixture of $$X$$ and $$Y$$ prepared in weight ratio $$2 : 3$$ should be:$$(P_{X}^{0} = 400\ torr, P_{Y}^{0} = 200\ torr)$$.

A
$$314\ torr$$

B
$$466.7\ torr$$

C
$$600\ torr$$

D
$$700\ torr$$

Solution

Let molecular mass of $$X = M$$ and molecular mass of $$Y = 2M$$
vapour pressure of $$X \, P_{x_0}^0 = 400 $$ torr
vapour pressure of $$Y \, P_y = 200$$ torr
Let weight of $$X = 2w$$
weight of $$Y = 3w$$
Mole fraction $$X_x = \dfrac{\dfrac{2w}{m}}{\dfrac{2w}{m} + \dfrac{3w}{2m}} = \dfrac{4}{7}$$
Mole fraction = $$Y_x = 1 - \dfrac{4}{7} = \dfrac{3}{7}$$
$$P = P_x^0 X_x + P_y^o Y_x$$
$$= 400 \times \dfrac{4}{7} + 200 \times \dfrac{3}{7} = \dfrac{2200}{7} $$
$$= 314 \, torr$$
Question 9
1 / -0

A nitrogen-hydrogen mixture initially in the molar ratio 1:3 reached equilibrium to form ammonia when 25% of the $$H_{2}$$ and $$N_{2}$$ had reacted. If the total pressure of the system was 21 atm, the partial pressure of ammonia at the equilibrium was:

A
4.5 atm

B
3.0 atm

C
2.0 atm

D
1.5 atm

Solution
Question 10
1 / -0

Two liquids X and Y are perfectly immiscible. If X and Y have molecular masses in ratio 1 : 2, the total vapour pressure of a mixture of X and Y prepared in weight ratio 2 : 3 should be ($$Px^0 = 400 torr, Py^0 = 200 torr$$)

A
314 torr

B
466.7 torr

C
600 torr

D
700 torr

Solution

Molecular mass of $$X = M$$
Molecular mass of $$Y = 2M$$
weight of $$X = 2W$$
weight of $$Y = 3w$$
no.of mole of $$X = \dfrac{2w}{M}$$
no. of mole of $$Y = \dfrac{3w}{2m}$$
mole fraction of $$X = Xx = \dfrac{\dfrac{2w}{m}}{\dfrac{2w}{M} + \dfrac{3w}{2M}}$$
$$X_x = \dfrac{4}{7}$$
Mole fraction of $$Y = Xy = \dfrac{3}{7}$$
$$P_x^o = 400 \, torr$$
$$P^o_y = 200 \, torr$$
$$P = \dfrac{4}{7} \times 400 + 200 \times \dfrac{3}{7}$$
$$= \dfrac{2200}{7} = 314 torr$$
option $$A$$ may be correct approximately