MHT CET · Chemistry · Solutions
A solution of non volatile solute has boiling point elevation 0.5 K . Calculate molality of solution \(\left[\mathrm{K}_{\mathrm{b}}=2.40 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right]\).
- A \(0.12 \mathrm{~mol} \mathrm{~kg}^{-1}\)
- B \(0.21 \mathrm{~mol} \mathrm{~kg}^{-1}\)
- C \(0.16 \mathrm{~mol} \mathrm{~kg}^{-1}\)
- D \(0.28 \mathrm{~mol} \mathrm{~kg}^{-1}\).
Answer & Solution
Correct Answer
(B) \(0.21 \mathrm{~mol} \mathrm{~kg}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \times \mathrm{m} \\ & 0.5 \mathrm{~K}=2.40 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \times \mathrm{m} \\ \therefore \quad & \mathrm{m}=\frac{0.5}{2.40}=0.21 \mathrm{~mol} \mathrm{~kg}^{-1}\end{aligned}\)
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