What is the freezing point of a 10% (by weight) solution of $C H_{3} O H$ in water?

$K_{f} = 1.86 K k g / m o l a l$

$- {6.45}^{\circ} C$

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${6.45}^{\circ} C$

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$- 10^{\circ} C$

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$90^{\circ} C$

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Solution

The correct option is A
$- {6.45}^{\circ} C$

Freezing point depression = $Δ T_{f} = K_{f} \times m$ .

1 kg solution has 0.100 kg $C H_{3} O H$ (solute) and 0.900 kg $H_{2} O$ (solvent)

GMM methanol = 32 g

The number of moles of methanol are: $100 g \times \frac{1 m o l}{32 g} = 3.12 m o l$

molality, $m = \frac{3.12 m o l}{0.900 k g} = 3.47 m$

Depression in freezing point $Δ T_{f} = K_{f} \times m = 1.86 \times 3.47 = {6.45}^{\circ} C$

$∴$ Freezing point of the solution

$= 0 - {6.45}^{\circ} C = - {6.45}^{\circ} C$

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