When 36 g of a non-volatile,non-electrolytic | vedclass

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(B) $\Delta T_f = K_f 	imes m$ $\Delta T_f = 0 - (-0.93) = 0.93 \ K$ $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{36 / M}

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$C_2H_4O_2$

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(B) $\Delta T_f = K_f 	imes m$ $\Delta T_f = 0 - (-0.93) = 0.93 \ K$ $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{36 / M}{1.2} = \frac{30}{M}$ $0.93 = 1.86 	imes \frac{30}{M}$ $M = \frac{1.86 	imes 30}{0.93} = 2 	imes 30 = 60 \ g \ mol^{-1}$ Empirical formula mass of $CH_2O = 12 + 2(1) + 16 = 30 \ g \ mol^{-1}$ $n = \frac{	ext{Molecular mass}}{	ext{Empirical formula mass}} = \frac{60}{30} = 2$ Molecular formula $= (CH_2O)_2 = C_2H_4O_2$

When $36 \ g$ of a non-volatile,non-electrolytic solute having the empirical formula $CH_2O$ is dissolved in $1.2 \ kg$ of water,the solution freezes at $-0.93 \ ^{\circ}C$. The molecular formula of the solute is ($K_f$ of water $= 1.86 \ K \ kg \ mol^{-1}$)

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