MHT CET · Chemistry · Solutions
Calculate the molality of solution of non volatile solute having depression in freezing point 0.93 K and cryoscopic constant of solvent \(1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\).
- A \(0.3 \mathrm{~mol} \mathrm{~kg}^{-1}\)
- B \(0.4 \mathrm{~mol} \mathrm{~kg}^{-1}\)
- C \(0.5 \mathrm{~mol} \mathrm{~kg}^{-1}\)
- D \(0.6 \mathrm{~mol} \mathrm{~kg}^{-1}\)
Answer & Solution
Correct Answer
(C) \(0.5 \mathrm{~mol} \mathrm{~kg}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\Delta \mathrm{T}_{\mathrm{f}} =\mathrm{K}_{\mathrm{f}} \mathrm{m} \)
\( \mathrm{m} =\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{f}}}=\frac{0.93 \mathrm{~K}}{1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}}=0.5 \mathrm{~mol} \mathrm{~kg}^{-1}\)
\( \mathrm{m} =\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{f}}}=\frac{0.93 \mathrm{~K}}{1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}}=0.5 \mathrm{~mol} \mathrm{~kg}^{-1}\)
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