MHT CET · Chemistry · Solutions
Calculate the cryoscopic constant of solvent when 2.5 gram solute is dissolved in 35 gram solvent lowers its freezing point by 3 K . (molar mass of solute is \(117 \mathrm{~g} \mathrm{~mol}^{-1}\) )
- A \(3.11 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)
- B \(3.56 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)
- C \(5.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)
- D \(4.91 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(D) \(4.91 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \Delta \mathrm{T}_{\mathrm{f}}=\frac{1000 \mathrm{~K}_{\mathrm{f}} \mathrm{~W}_2}{\mathrm{M}_2 \mathrm{~W}_1} \\
& \mathrm{~K}_{\mathrm{f}}=\frac{\Delta \mathrm{T}_{\mathrm{f}} \times \mathrm{M}_2 \times \mathrm{W}_{\mathrm{i}}}{1000 \times \mathrm{W}_2} \\
& =\frac{3 \mathrm{~K} \times 117 \mathrm{~g} \mathrm{~mol}^{-1} \times 35 \mathrm{~g}}{1000 \mathrm{gkg}^{-1} \times 2.5 \mathrm{~g}} \\
& =4.91 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}
\end{aligned}\)
& \Delta \mathrm{T}_{\mathrm{f}}=\frac{1000 \mathrm{~K}_{\mathrm{f}} \mathrm{~W}_2}{\mathrm{M}_2 \mathrm{~W}_1} \\
& \mathrm{~K}_{\mathrm{f}}=\frac{\Delta \mathrm{T}_{\mathrm{f}} \times \mathrm{M}_2 \times \mathrm{W}_{\mathrm{i}}}{1000 \times \mathrm{W}_2} \\
& =\frac{3 \mathrm{~K} \times 117 \mathrm{~g} \mathrm{~mol}^{-1} \times 35 \mathrm{~g}}{1000 \mathrm{gkg}^{-1} \times 2.5 \mathrm{~g}} \\
& =4.91 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}
\end{aligned}\)
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