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(D) The depression in freezing point is given by $\Delta T_f = T_f^o - T_f = 0 - (-0.465) = 0.465 \ K$. Using the formula for molar mass $M$ of the so

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$C_6H_{12}O_6$

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(D) The depression in freezing point is given by $\Delta T_f = T_f^o - T_f = 0 - (-0.465) = 0.465 \ K$. Using the formula for molar mass $M$ of the solute: $M = \frac{1000 	imes K_f 	imes w_2}{\Delta T_f 	imes w_1} = \frac{1000 	imes 1.86 	imes 1.8}{0.465 	imes 40} = 180 \ g \ mol^{-1}$. The empirical formula mass of $CH_2O$ is $12 + (2 	imes 1) + 16 = 30 \ g \ mol^{-1}$. Calculating the value of $n$: $n = \frac{	ext{Molecular mass}}{	ext{Empirical formula mass}} = \frac{180}{30} = 6$. Therefore,the molecular formula is $(CH_2O)_6 = C_6H_{12}O_6$.

$A$ solution containing $1.8 \ g$ of a compound (empirical formula $CH_2O$) in $40 \ g$ of water is observed to freeze at $-0.465 \ ^oC$. The molecular formula of the compound is ($K_f$ of water $= 1.86 \ K \ kg \ mol^{-1}$)

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