Question

Equal weight of a solute are dissolved in equal weight of two solvents $A\&B$ and formed a very dilute solution. The relative lowering of vapour pressure for solution B has twice the relative lowering of vapour pressure for solution A. If $M_A$ and $M_B$ are molecular weights of solvents $A\&B$ respectively, then:

• A. $M_B=2M_A$
• B. $M_A=4M_B$
• C. $M_A=2M_B$
• D. $M_A=M_B$

Answer 1

Answer: (A)

${\left(\frac{P^0-P_s}{P^0}\right)}_A=X_{\text{solute }}$ in solvent $A(\because$ solute is same $)$

${\left(\frac{P^0-P_s}{P^0}\right)}_B=2{\left(\frac{P^0-P_s}{P^0}\right)}_A$

${\left(\frac{n_{\text{solute }}}{n_{\text{solute }}+n_{\text{solvent }}}\right)}_{\text{in }B}=2{\left(\frac{n_{\text{solute }}}{n_{\text{solute }}+n_{\text{solvent }}}\right)}_{\text{in }A}$

$X_{\text{solute }}$ in solvent $B=2X_{\text{solute }}$ in solvent $A$

Since solution is very dilute-

$n_{\text{solute }}+n_{\text{solvent }}\approx n_{\text{solvent }}$

$\therefore {\left(\frac{n_{\text{solute }}}{n_{\text{solvent }}}\right)}_{\text{in }B}=2{\left(\frac{n_{\text{solute }}}{n_{\text{solvent }}}\right)}_{\text{in }A}$

Now, no. of moles of solute in solvent $A\&B$ are same,

$\therefore \frac{1}{n_{\text{solvent in B }}}=\frac{1}{n_{\text{solvent in A }}}$

$\frac{1}{W_{\text{solvent }}}\times M_{{\text{solvent }}_B}=\frac{1}{w_{\text{solvent }}}\times M_{\text{solvent }}$

(gives weights of solvents are also equal)

$\therefore M_B=2M_A$