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The freezing point of a solution prepared from $1.25 \mathrm{gm}$ of a non-electrolyte and $20 \mathrm{gm}$ of water is $271.9 \mathrm{~K}$. If molar depression constant is $1.86 \mathrm{~K} \mathrm{~mole}^{-1}$, then molar mass of the solute will be
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The correct answer is:
105.7
Molar mass $=\frac{K_f \times 1000 \times W}{\Delta T_f \times W}=\frac{1.86 \times 105.7 \times 1.25}{20 \times 1.1}$
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