MHT CET · Chemistry · Solutions
\(5 \mathrm{~g}\) sucrose (molar mass \(=342\) ) is dissolved in \(100 \mathrm{~g}\) of solvent, decreases the freezing point by \(2.15 \mathrm{~K}\). What is cryoscopic constant of solvent?
- A \(14.7 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)
- B \(2.15 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)
- C \(4.30 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)
- D \(7.35 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(A) \(14.7 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f} \mathrm{m}} \\ & \mathrm{m}=\frac{\text { moles of solute }}{\text { massof solvent }} \times 1000 \\ & \mathrm{~m}=\frac{5 / 342}{100} \times 1000=\frac{50}{342} \\ & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{kf} \cdot \mathrm{m} \\ & 2.15=\mathrm{kf} \cdot \frac{50}{342} \\ & \mathrm{kf}=\frac{2.15 \times 342}{50}=14.7 \mathrm{~K} \mathrm{~kg} \cdot \mathrm{mol}^{-1}\end{aligned}\)
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