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. $M_{B}=2 M_{A}$
• B. $M_{A}=4 M_{B}$
• C. $M_{A}=2 M_{B}$
• 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 $B=2 X_{\text {solute }}$ 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)

$\therefore M _{ B }=2 M_A$