KCET · Chemistry · Solutions
\(3 \mathrm{~g}\) of urea is dissolved in \(45 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\). The relative lowering in vapour pressure is
- A \(0.05\)
- B \(0.04\)
- C \(0.02\)
- D \(0.01\)
Answer & Solution
Correct Answer
(C) \(0.02\)
Step-by-step Solution
Detailed explanation
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
\(n_{2}=\frac{w_{2}}{M_{2}}=\frac{3}{60}=\frac{1}{20}=0.05\)
\(n_{1}=\frac{w_{1}}{M_{1}}=\frac{45}{18}=2.5\)
\(\therefore \quad \frac{p^{\circ}-p_{s}}{p^{\circ}}=\frac{n_{2}}{n_{1}+n_{2}}=\frac{0.05}{2.5+0.05}=0.02\)
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