COMEDK · Chemistry · 17. Solutions
Given that the freezing point of benzene is \(5.48^{\circ} \mathrm{C}\) and its \(\mathrm{K}_{\mathrm{f}}\) value is \(5.12{ }^{\circ} \mathrm{C} / \mathrm{m}\). What would be the freezing point of a solution of \(20 \mathrm{~g}\) of propane in \(400 \mathrm{~g}\) of benzene?
- A \(-0.2{ }^{\circ} \mathrm{C}\)
- B \(-0.34^{\circ} \mathrm{C}\)
- C \(-0.17^{\circ} \mathrm{C}\)
- D \(-5.8^{\circ} \mathrm{C}\)
Answer & Solution
Correct Answer
(B) \(-0.34^{\circ} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
The molar mass of propane (\(C_{3}H_{8}\)) is \(3 \times 12 + 8 \times 1 = 44 \text{ g/mol}\). The number of moles of propane is \(n = \dfrac{20 \text{ g}}{44 \text{ g/mol}} = \dfrac{5}{11} \text{ mol} \approx 0.4545 \text{ mol}\). The mass of the solvent (benzene) is…
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