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$6 \mathrm{~g}$ of a non-volatile, non-electrolyte $X$ dissolved in $100 \mathrm{~g}$ of water freezes at $-0.93^{\circ} \mathrm{C}$. The molar mass of $X$ in $\mathrm{g} \mathrm{mol}^{-1}$ is $\left(K_f\right.$ of $\left.\mathrm{H}_2 \mathrm{O}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$
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The correct answer is:
$120$
$\Delta \mathrm{T}_f=\mathrm{K}_f m$
$0.93=\frac{1.86 \times 6 \times 1000}{M \times 100} \Rightarrow \mathrm{M} \approx 120$
$0.93=\frac{1.86 \times 6 \times 1000}{M \times 100} \Rightarrow \mathrm{M} \approx 120$
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