A solution containing $30 \mathrm{~gms}$ of non-volatile solute in exactly $90 \mathrm{~gm}$ water has a…

A solution containing $30 \mathrm{~gms}$ of non-volatile solute in exactly $90 \mathrm{~gm}$ water has a vapour pressure of $21.85 \mathrm{~mm} \mathrm{~Hg}$ at $25^{\circ} \mathrm{C}$. Further $18 \mathrm{~gms}$ of water is then added to the solution. The resulting solution has a vapour pressure of $22.15 \mathrm{~mm} \mathrm{~Hg}$ at $25^{\circ} \mathrm{C}$. Calculate the molecular weight of the solute
  1. $74.2$
  2. $75.6$
  3. $67.83$
  4. $78.7$

Solution

We have,
$\frac{p^{\circ}-21.85}{21.85}=\frac{30 \times 18}{90 \times m}$ for $\mathrm{I}$ case......(i)
wt. of solvent $=90+18=108 \mathrm{~gm}$
$\frac{p^{\circ}-22.15}{22.15}=\frac{30 \times 18}{108 \times m}$, for $\mathrm{II}$ case......(ii)
By eq. $(1)$
$p^{\circ} m-21.85 m=21.85 \times 6=131.1$
By eq. $(2)$
$p^{\circ} m-22.15 m=22.15 \times 5=110.75$
$0.30 m=20.35$
$m=\frac{20.35}{0.30}=67.83$ .

Asked in: JEE-TOPICTESTS-CHEMISTRY

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