Question:
The relative lowering of vapour pressure of an aqueous solution containing a non-volatile solute is 0.0125. The molality of the solution is
The relative lowering of vapour pressure of an aqueous solution containing a non-volatile solute is 0.0125. The molality of the solution is
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For aqueous dilute solutions, \(55.56\) moles of water are present in \(1\,kg\) of solvent.
Updated On: Jun 18, 2026
- 0.65 m
- 0.35 m
- 0.70 m
- 0.30 m
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The Correct Option is C
Solution and Explanation
Concept:
For dilute solutions,
\[
\frac{p^0-p}{p^0}
=
\frac{n_2}{n_1+n_2}
\]
and approximately
\[
\frac{p^0-p}{p^0}
\approx
\frac{n_2}{n_1}
\]
where \(n_2\) is moles of solute and \(n_1\) is moles of solvent.
Step 1: Use the given relative lowering.
\[ \frac{p^0-p}{p^0}=0.0125 \] Take \[ 1000g \] of water. Then \[ n_1=\frac{1000}{18}=55.56 \]
Step 2: Calculate moles of solute.
\[ 0.0125=\frac{n_2}{55.56} \] \[ n_2=0.6945 \]
Step 3: Determine molality.
Molality \[ m=\frac{n_2}{1} \] \[ m\approx0.695 \] \[ \boxed{m=0.70} \]
Step 1: Use the given relative lowering.
\[ \frac{p^0-p}{p^0}=0.0125 \] Take \[ 1000g \] of water. Then \[ n_1=\frac{1000}{18}=55.56 \]
Step 2: Calculate moles of solute.
\[ 0.0125=\frac{n_2}{55.56} \] \[ n_2=0.6945 \]
Step 3: Determine molality.
Molality \[ m=\frac{n_2}{1} \] \[ m\approx0.695 \] \[ \boxed{m=0.70} \]
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