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Q.

At 1000C the vapour pressure of a solution of 6.5 gm of a solute in 100 gm of water is 732 mm. If Kb = 0.52, the boiling point of this solution will be

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a

103oC

b

101oC

c

102oC

d

1000C

answer is C.

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Detailed Solution

At 100oc

             

P_{H_2O}^o=760mm(1atm)\\P_{sol.}=732mm\\W_{solute}=6.5g\\W_{H_2O}=100g

              GMW of H2O=18g

           Kb(H2O)= 0.52KKgmole-1

(I)     Assuming the solution is very dilute

      

\frac{P^o-P_s}{P^o}X\frac{1000}{GMW\;of\;solvent}=molality
m=\frac{760-732}{760}X\frac{1000}{18}

m=2.05m

(II)

           

\Delta T_b=K_bm=0.52(2.05)=1.06K

              Tb(solution)-100=1.06

             Tb(solution) = 101.060C

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