Question

If mole fraction of solvent is ' \( s \) ' and lowering of vapour pressure is ' \( p \) ' mm of Hg then, what will be vapour pressure of pure solvent (Po)? (assume solute is nonvolatile and nonelectrolyte) 

(1) \( p^{\circ}=s p \) 

(2) \( p ^{\circ}= p ^{\prime} s \) 

(3) \( p^{\circ}=s / p \) 

(4) \( p^{0}=p /(1-s) \)

Answer 1

Correct option is (4) P^o = \(\frac{P}{1-S}\) 

As we know, relative lowering of vapour pressure is given by--

\(\frac{P^o-P_1}{P^o}=x\)----(1)

where,

P^o - P₁ = lowering in vapour pressure

P^o = Vapour pressure of pure solvent

x = mole fraction of solute

we have given,

lowering of vapour pressure = P^o - P₁ = P

mole fraction of solvent  = 'S'

∴ mole fraction of solute = 1 - S

putting these values in equation (1)

\(\frac{P}{P^o}=1-S \) 

P^o = \(\frac{P}{1-S}\)

P^o = \(\frac{P}{1-S}\) 

Therefore, vapour pressure of pure solvent will be

P^o = \(\frac{P}{1-S}\)