AP EAMCET · Chemistry · Solutions
What is the depression of freezing point, when mole fraction of non-electrolyte solute in aqueous solution is \(0.01 ?\left(\mathrm{~K}_{\mathrm{f}}\right.\) of \(\left.\mathrm{H}_2 \mathrm{O}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)\)
- A \(1.246 \mathrm{~K}\)
- B \(1.380 \mathrm{~K}\)
- C \(1.528 \mathrm{~K}\)
- D \(1.043 \mathrm{~K}\)
Answer & Solution
Correct Answer
(D) \(1.043 \mathrm{~K}\)
Step-by-step Solution
Detailed explanation
\(\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{m} \times \mathrm{K}_{\mathrm{f}}=\frac{\mathrm{n}_2}{\mathrm{~m}_1} \times \mathrm{K}_{\mathrm{f}}\) We have, \(\mathrm{x}_2=\frac{\mathrm{n}_2}{\mathrm{n}_1+\mathrm{n}_2} \approx \frac{\mathrm{n}_2}{\mathrm{n}_1}=0.01\) Taking…
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