Search any question & find its solution
Question:
Answered & Verified by Expert
$3 \mathrm{~g}$ of urea is dissolved in $45 \mathrm{~g}$ of $\mathrm{H}_{2} \mathrm{O}$. The relative lowering in vapour pressure is
Options:
Solution:
2233 Upvotes
Verified Answer
The correct answer is:
$0.02$
Relative lowering of vapour pressure
$\frac{p^{\circ}-p_{s}}{p^{\circ}}=\frac{n_{2}}{n_{1}+n_{2}}$
where $p^{\circ}=$ vapour pressure of the pure solvent.
$p_{s}=$ vapour pressure of the solution
$n_{2}=$ number of moles of the solute
$n_{1}=$ number of moles of the solvent
$n_{2}=\frac{w_{2}}{M_{2}}=\frac{3}{60}=\frac{1}{20}=0.05$
$n_{1}=\frac{w_{1}}{M_{1}}=\frac{45}{18}=2.5$
$\therefore \quad \frac{p^{\circ}-p_{s}}{p^{\circ}}=\frac{n_{2}}{n_{1}+n_{2}}=\frac{0.05}{2.5+0.05}=0.02$
$\frac{p^{\circ}-p_{s}}{p^{\circ}}=\frac{n_{2}}{n_{1}+n_{2}}$
where $p^{\circ}=$ vapour pressure of the pure solvent.
$p_{s}=$ vapour pressure of the solution
$n_{2}=$ number of moles of the solute
$n_{1}=$ number of moles of the solvent
$n_{2}=\frac{w_{2}}{M_{2}}=\frac{3}{60}=\frac{1}{20}=0.05$
$n_{1}=\frac{w_{1}}{M_{1}}=\frac{45}{18}=2.5$
$\therefore \quad \frac{p^{\circ}-p_{s}}{p^{\circ}}=\frac{n_{2}}{n_{1}+n_{2}}=\frac{0.05}{2.5+0.05}=0.02$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.