MHT CET · Chemistry · Solutions
What is the value of \(\mathrm{K}_{\mathrm{f}}\) if \(30 \mathrm{~g}\) urea (molar mass 60 ) dissolved in \(0.5 \mathrm{dm}^{3}\) of water decreases freezing point by \(0 \cdot 15^{\circ} \mathrm{C}\) ?
- A \(0 \cdot 15 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)
- B \(0.030 \mathrm{Kkg} \mathrm{mol}^{-1}\)
- C \(0.30 \mathrm{Kkg} \mathrm{mol}^{-1}\)
- D \(0.015 \mathrm{Kkg} \mathrm{mol}^{-1}\)
Answer & Solution
Correct Answer
(A) \(0 \cdot 15 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
Mass of water \(=0.5 \mathrm{dm}^{3}=0.5 \mathrm{~kg}\) Moles of urea \(=\frac{30}{60}=0.5\) mole
Molality of urea \(=\frac{0.5 \mathrm{~mol}}{0.5 \mathrm{~kg}}=1 \mathrm{~mol} / \mathrm{kg}\)
\(\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \times \mathrm{m}\)
\(\mathrm{K}_{\mathrm{f}}=\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\mathrm{m}}=\frac{0.15 \mathrm{~K}}{1 \mathrm{~mol} / \mathrm{kg}}=0.15 \mathrm{Kmol}^{-1} \mathrm{~kg}\)
Molality of urea \(=\frac{0.5 \mathrm{~mol}}{0.5 \mathrm{~kg}}=1 \mathrm{~mol} / \mathrm{kg}\)
\(\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \times \mathrm{m}\)
\(\mathrm{K}_{\mathrm{f}}=\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\mathrm{m}}=\frac{0.15 \mathrm{~K}}{1 \mathrm{~mol} / \mathrm{kg}}=0.15 \mathrm{Kmol}^{-1} \mathrm{~kg}\)
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