MHT CET · Chemistry · Solutions
\(0.5\) molal aqueous solution of a weak acid \((\mathrm{HX})\) is \(20 \%\). ionised. If \(\mathrm{K}_{\mathrm{f}}\) of water is \(1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\), the lowering in freezing point of solution is
- A \(-0 \cdot 56 \mathrm{~K}\)
- B \(-1 \cdot 12 \mathrm{~K}\)
- C \(1 \cdot 12 \mathrm{~K}\)
- D \(0.56 \mathrm{~K}\)
Answer & Solution
Correct Answer
(C) \(1 \cdot 12 \mathrm{~K}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{HX} \rightleftharpoons \mathrm{H}^{+}+\mathrm{X}^{-}\)
For dissociation of an electrolyte,
\(\alpha=\frac{i-1}{n-1} \quad \therefore 0.2=\frac{i-1}{2-1}\)
\(\therefore \mathrm{i}-1=0.2\)
\(\therefore \mathrm{i}=0.2+1=1.2\)
Now, \(\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{k}_{\mathrm{f}} \mathrm{m}=1.2 \times 1.86 \times 0.5=1.12 \mathrm{~K}\)
For dissociation of an electrolyte,
\(\alpha=\frac{i-1}{n-1} \quad \therefore 0.2=\frac{i-1}{2-1}\)
\(\therefore \mathrm{i}-1=0.2\)
\(\therefore \mathrm{i}=0.2+1=1.2\)
Now, \(\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{k}_{\mathrm{f}} \mathrm{m}=1.2 \times 1.86 \times 0.5=1.12 \mathrm{~K}\)
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