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
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:
the correct answer is c) 90 amu