At 100^oC the vapour pressure of a solution of 6.5 g of a non-volatile solute in 100 g water is 732 mm. If K_b=0.52 , the boiling point of this…

Question

At 100^oC the vapour pressure of a solution of 6.5 g of a non-volatile solute in 100 g water is 732 mm. If K_b=0.52 , the boiling point of this solution will be:

MARKS Content Team · Accepted Answer

Correct answer: (A) 101^oC. At 100^oC (boiling point) Vapour pressure of water P^o=P_atm=760 mm ∴(P^o-P_s)/(P^o)=X_solute ⇒(760-732)/760=n_solute/n_solvent{m=molar mass of solute} ⇒28/732=(6.5/m)/(100/18 ) ⇒m=(6.5×18×732)/(28×100)≈30.6 Now, ∆T_b=K_b× molality =0.52×(6.5/30.6)/0.1=1.1°C ∴ Boiling point of solution =100+1.1=101.1^oC ≃101°C

At $100^{o} C$ the vapour pressure of a solution of 6.5 g of a non-volatile solute in 100 g water is 732 mm. If $K_{b} = 0.52$ , the boiling point of this solution will be:

Solution

Relevant Questions

At 100oC the vapour pressure of a solution of 6.5 g of a non-volatile solute in 100 g water is 732 mm. If Kb=0.52 , the boiling point of this solution will be:

Solution

Relevant Questions

At $100^{o} C$ the vapour pressure of a solution of 6.5 g of a non-volatile solute in 100 g water is 732 mm. If $K_{b} = 0.52$ , the boiling point of this solution will be: