MHT CET · Chemistry · Solutions
Calculate the molar mass of non volatile solute when 5 g of it is dissolved in 50 g solvent, boils at \(119.6^{\circ} \mathrm{C}\).
\(\left[\mathrm{K}_{\mathrm{b}}=3.2 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right.\), boiling point of pure solvent \(\left.=118^{\circ} \mathrm{C}\right]\).
- A \(180 \mathrm{~g} \mathrm{~mol}^{-1}\)
- B \(210 \mathrm{~g} \mathrm{~mol}^{-1}\)
- C \(200 \mathrm{~g} \mathrm{~mol}^{-1}\)
- D \(190 \mathrm{~g} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(C) \(200 \mathrm{~g} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{T}_{\mathrm{b}}-\mathrm{T}_{\mathrm{b}}^0=119.6-118=1.6^{\circ} \mathrm{C}=\) \(1.6 \mathrm{~K} \)
\( \Delta \mathrm{~T}_{\mathrm{b}}=\frac{1000 \mathrm{~K}_{\mathrm{b}} \mathrm{W}_2}{\mathrm{M}_2 \mathrm{~W}_1} \)
\( \mathrm{M}_2=\frac{1000 \mathrm{~K}_{\mathrm{b}} \mathrm{W}_2}{\Delta \mathrm{~T}_{\mathrm{b}} \mathrm{W}_1}=\) \(\frac{1000 \times 3.2 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}{ }^{-1} \times 5 \mathrm{~g}}{1.6 \mathrm{~K} \times 50 \mathrm{~g}} \)
\(=200 \mathrm{~g} \mathrm{~mol}^{-1}\)
\( \Delta \mathrm{~T}_{\mathrm{b}}=\frac{1000 \mathrm{~K}_{\mathrm{b}} \mathrm{W}_2}{\mathrm{M}_2 \mathrm{~W}_1} \)
\( \mathrm{M}_2=\frac{1000 \mathrm{~K}_{\mathrm{b}} \mathrm{W}_2}{\Delta \mathrm{~T}_{\mathrm{b}} \mathrm{W}_1}=\) \(\frac{1000 \times 3.2 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}{ }^{-1} \times 5 \mathrm{~g}}{1.6 \mathrm{~K} \times 50 \mathrm{~g}} \)
\(=200 \mathrm{~g} \mathrm{~mol}^{-1}\)
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