A solution of 1.25 g of a non - electrolyte in 20 g of water freezes at 271.94 K. If $K_{f}$ = 1.86 K molality -1 then the molecular wt. of the solute is

207.8 g/ mol

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179.79 g/ mol

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109.6 g / mol

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96.01 g / mol

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Solution

The correct option is D 109.6 g / mol
Freezing point of solution $= 271.94 K$
Freezing point of water $= 273.0 K$
$Δ T = (273 - 271.94) = 1.06 K$
We know that $m = \frac{1000 K_{f} \times w}{W Δ T}$
Given , $K_{f} = 1.86 / K / k g m o l^{- 1}, w = 1.25 g, W = 20 g$ and $Δ = 1.06 K$
$m = \frac{1000 \times 1.86 \times 1.25}{20 \times 1.06} = 109.66 g / m o l$

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A solution of 1.25g of non - electrolyte in 20g of water freezes at 271.94 K. If K_f = 1.86 K kg mol^-1, then the molecular weight of the solute will be:

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