MHT CET · Chemistry · Solutions
If \(6 \mathrm{~g}\) of solute dissolved in \(100 \mathrm{~g}\) of water lowers the freezing point by \(0.93 \mathrm{~K}\). What is molar mass of solute?
\(
\left(\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)
\)
- A \(120 \mathrm{~g} \mathrm{~mol}^{-1}\)
- B \(60 \mathrm{~g} \mathrm{~mol}^{-1}\)
- C \(90 \mathrm{~g} \mathrm{~mol}^{-1}\)
- D \(180 \mathrm{~g} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(A) \(120 \mathrm{~g} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{m}=\frac{\text { moles of solute }}{\text { mass of solvent }} \times 1000 \)
\( =\frac{6 / \mathrm{M}}{100} \times 1000=\frac{60}{\mathrm{M}} \)
\( \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \cdot \mathrm{m} \)
\( 0.93=1.86 \times \frac{60}{\mathrm{M}} \Rightarrow \mathrm{M}=\frac{1.86 \times 60}{0.93}\) \(=120 \mathrm{~g} \mathrm{~mol}^{-1}\)
\( =\frac{6 / \mathrm{M}}{100} \times 1000=\frac{60}{\mathrm{M}} \)
\( \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \cdot \mathrm{m} \)
\( 0.93=1.86 \times \frac{60}{\mathrm{M}} \Rightarrow \mathrm{M}=\frac{1.86 \times 60}{0.93}\) \(=120 \mathrm{~g} \mathrm{~mol}^{-1}\)
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