Equal masses of a solute are dissolved in equal amount of two solvents $A$ and $B$, respective molecular…

Equal masses of a solute are dissolved in equal amount of two solvents $A$ and $B$, respective molecular masses being $M_{A}$ and $M_{B}$. The relative lowering of vapour pressure of solution in solvent $A$ is twice that of the solution in solvent $B$. If the solutions are dilute, $M_{A}$ and $M_{B}$ are related as
  1. $M_{A}=M_{B}$
  2. $2 M_{A}=M_{B}$
  3. $M_{A}=2 M_{B}$
  4. $M_{A}=4 M_{B}$

Solution

For dilute solution,
$\frac{\Delta \mathrm{P}}{\mathrm{P}^{\circ}} \Rightarrow \frac{\mathrm{n}_{\text {solute }}}{\mathrm{n}_{\text {solvent }}}$ For solution in $\mathrm{A}$,
$\frac{\Delta \mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{A}}^{\circ}}=\frac{\mathrm{W} / \mathrm{M}}{\mathrm{W}_{\mathrm{A}} / \mathrm{M}_{\mathrm{A}}}=\frac{\mathrm{W}}{\mathrm{M}} \times \frac{\mathrm{M}_{\mathrm{A}}}{\mathrm{W}_{\mathrm{A}}}$
For solution in $\mathrm{B}, \frac{\Delta \mathrm{P}_{\mathrm{B}}}{\mathrm{P}_{\mathrm{A}}^{\circ}}=\frac{\mathrm{W}}{\mathrm{M}} \times \frac{\mathrm{M}_{\mathrm{B}}}{\mathrm{W}_{\mathrm{B}}} \quad \ldots$..(ii)
From (i) and (ii), $\frac{\Delta \mathrm{P}_{\mathrm{A}} / \mathrm{P}_{\mathrm{A}}^{\circ}}{\Delta \mathrm{P}_{\mathrm{B}} / \mathrm{P}_{\mathrm{B}}^{\circ}}=2=\frac{\mathrm{M}_{\mathrm{A}} \mathrm{W}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{B}} \mathrm{W}_{\mathrm{A}}}=\frac{\mathrm{M}_{\mathrm{A}}}{\mathrm{M}_{\mathrm{B}}}\left(\mathrm{W}_{\mathrm{A}}=\mathrm{W}_{\mathrm{B}}ight)$
$\mathrm{M}_{\mathrm{A}}=2 \mathrm{M}_{\mathrm{B}}$ ,

Asked in: JEE-TOPICTESTS-CHEMISTRY

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