Question

The mass of a non-volatile, non-electrolyte solute (molar mass =50 g mol 1) needed to be dissolved in 114 g octane to reduce its vapour pressure to 75%, is

Answer 1

Hii dear,

● Answer -
W2 = 16.67 g

● Explaination -
# Given -
W1 = 111 g
W2 = ?
M1 = 114 g
M2 = 50 g

# Solution -
Relative lowering of vapor pressure is calculated by -
∆P/P° = 100 - 75 = 25 %

But we know, relative lowering of vapor pressure is equal to mole fraction of solute.
∆P/P° = X2 = n2/(n1+n2)
25/100 = n2 / (1 + n2)
n2 = 0.33 mol

Weight of solute is -
W2 = n2 × M2
W2 = 0.33 × 50
W2 = 16.67 g

Weight of the solute is 16.67 g.


Hope this helps you..