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JEE Advanced levelOne or more correct optionsDifficulty 3 of 3so070

$0.40 mol$ of glucose is dissolved in $500 g$ of water ( $K_{b} = 0.52$ and $K_{f} = 1.86 K kg mo l^{- 1}$ ; normal boiling point $100.00^{\circ} C$ ). Which of the following is/are correct?

AThe freezing point is $- 1.49^{\circ} C$ (two decimals).
BThe ratio $Δ T_{f} /Δ T_{b}$ is $3.58$ .
CThe boiling point is $100.42^{\circ} C$ (two decimals).
DThe boiling point is $100.83^{\circ} C$ (two decimals).

Answer: A, B, C

Molality

= 0.40/0.500 = 0.80 mol k g^{- 1}

.

Δ T_{b} = 0.52 \times 0.80 = 0.416 K

, so the boiling point is

100.42^{\circ} C

;

Δ T_{f} = 1.86 \times 0.80 = 1.488 K

, so

T_{f} = - 1.49^{\circ} C

. The ratio is

K_{f} / K_{b} = 3.58

. The value

100.83^{\circ} C

would need the molality doubled, so it is wrong.

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