Chemistry, asked by Anonymous, 1 month ago

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​

Answers

Answered by ItzCutePrince1946
9

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

Answered by gaikwaddipti743
1

Answer:

the correct answer is c) 90 amu

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