Question

The vapour pressure of pure A is 10 torr and at

the same temperature when 1 g of B is dissolved in

20 gm of A, its vapour pressure is reduced to 9.0

torr. If the molecular mass of A is 200 amu, then

the molecular mass of B is :

a)100amu
b)75amu
c)90amu
d)120amu​

Answer 1

Answer:

• The correct option is: c) 90 amu

Explanation:

Given that:

• Vapour pressure of pure A = 10 torr
• Vapour pressure of solution = 9 torr
• Weight of A = 20 gram
• Weight of B = 1 gram
• Molecular mass of A = 200 amu

To Find:

• The molecular mass of B.

Let us assume:

• The molecular mass of B be x.

Formula used:

• n = w/m.w.
• (P° - P)/P = n₂/(n₁ + n₂)

Where,

• n = Mole
• w = Weight
• m.w. = Molecular weight
• P° = Vapour pressure of pure liquid
• P = Vapour pressure of solution
• n₁ = Mole of A
• n₂ = Mole of B

Finding the mole of A:

We have,

• w = 20 gram
• m.w. = 200 amu

• n₁ = 20/200
• n₁ = 1/10

Finding the mole of B:

We have,

• w = 1 gram
• m.w. = x

• n₂ = 1/x

Finding the value of x:

We have,

• n₁ = 1/10
• n₂ = 1/x
• P° = 10
• P = 9

⟶ (P° - P)/P = n₂/(n₁ + n₂)

⟶ (10 - 9)/10 = (1/x)/(1/10 + 1/x)

⟶ 1/10 = (1/x)/(1/10 + 1/x)

Cross multiplication.

⟶ 1/10 + 1/x = 10/x

⟶ (x + 10)/10x = 10/x

Cross multiplication.

⟶ x(x + 10) = 10(10x)

⟶ x² + 10x = 100x

⟶ x² = 100x - 10x

⟶ x² = 90x

⟶ x•x = 90•x

Cancelling x both sides.

⟶ x = 90

∴ The molecular mass of B = 90 amu

Answer 2

Given :-

The  vapour pressure of pure A is 10 torr and at the same temperature when 1 g of B is dissolved in 20 gm of A, its vapour pressure is reduced to 9.0 torr. If the molecular mass  of A is 200 amu,

To Find :-

Molecular mass of B

Solution :-

$\sf \dfrac{10-9}{10} = \dfrac{\dfrac{1}{x}}{\dfrac{20}{200}+\dfrac{1}{x}}$

$\sf \dfrac{1}{10} = \dfrac{\dfrac{1}{x}}{\dfrac{20}{200}+\dfrac{1}{x}}$

$\sf \sf \dfrac{1}{10} = \dfrac{\dfrac{1}{x}}{\dfrac{1}{10}+\dfrac{1}{x}}$

$\sf \dfrac{1}{10}\bigg(\dfrac{1}{10} + \dfrac{1}{x}\bigg) = \dfrac{1}{x}$

$\sf \dfrac{1}{100} + \dfrac{1}{10x} = \dfrac{1}{x}$

$\sf \dfrac{1}{100} = \dfrac{1}{x}- \dfrac{1}{10x}$

$\sf \dfrac{1}{100} = \dfrac{10x-x}{10}$

$\sf 10= 100(9x)$

$\sf x = \dfrac{900}{10} = 90$