Question

If the value of lowering in vapour pressure of a liquid is 100 mm Hg when a non -elctrolyte solute is added to it. What will be the value of elevation in boiling point if vapour pressure of pure liquid is 700 mm Hg and $K_b=10Kkg/mol$? (Molar mass of solvent = 167 g)

• A. 5 K
• B. 10 K
• C. 15 K
• D. 20 K

Answer 1

The correct option is B 10 K

Depression in vapour pressure is caused due to presence of non-volatile particles. The drop can be calculated by:
Drop in Vapour pressure = Vapour pressure of pure liquid $\times$ mole fraction
Or, 100 mm = 700 mm $\times$ mole fraction
Or, Mole fraction = $\frac{1}{7}$
Moles of Solute 1
Moles of Solvent 6
Molality of Solution = $\frac{Molesofsolute}{kgofSolvent}=\frac{1}{0.167\times 6}=\frac{1}{1}=1$
Elevation in Boiling point is given by:
Elevation in Boiling Point = $Kb\times molality$
$=10K\times 1=10K$.