Question

The vapour pressure of a solution of a non volatile electrolyte(A) in a solvent(B) is 95% of vapour pressure of solvent at the same temp. If MB = 0.3 MA where MB & MA are molecular wts. of B & A resp, what is the weight ratio of solvent & solute?

Answer 1

we know ,
relative lowering of vapor pressure is the ratio of mole of solute to the mole of solvent.
e.g., $\bold{\frac{\Delta{P}}{P_0}=\frac{n_{solute}}{n_{solvent}}}$
Or, ∆P/P₀ = wM/Wm
Where w is the weight of solute
M is the molecular weight of solvent
W is the weight of solute
m is the molecular weight of solute
∆P is the lowering of vapor pressure
And P₀ is the initial vapor pressure
Given,
MB = 0.3MA
Let the intial vapor pressure ,P₀ = 100 Pa
Then, lowering of vapor pressure ,∆P = (100 - 95) Pa = 5 Pa
now,
5/100 = w × 0.3MA/W × MA
⇒1/20 = 0.3w/W
⇒w/W = 1/6

Hence , weight ratio of solute and solvent is 1/6