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The freezing point of a solution containing 36g of a compound (empirical formula : $CH_2O$) in 1.20 kg of water is found to be $-0.93^{\large\circ}C$.The molecular formula of the solute is

$\begin{array}{1 1}CH_2O\\C_2H_4O_2\\C_3H_6O_3\\C_4H_8O_4\end{array}$

Can you answer this question?

Answer : $C_2H_4O_2$
$-\Delta T_f=K_fm$
$\Rightarrow 0.93K=(1.86K\; Kg\;mol^{-1})\big(\large\frac{36g/M}{1.20kg}\big)$
This gives $M=\large\frac{1.86\times 36}{0.93\times 1.20}$$gmol^{-1} n=\large\frac{\text{Molar mass}}{\text{Empirical molar mass}} \;\;=\large\frac{60gmol^{-1}}{30gmol^{-1}}$$=2$
Hence,molecular formula is $C_2H_4O_2$
answered Jul 23, 2014