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

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(B) Given: Mass of solute $= 36 \ g$,Mass of solvent $= 1.2 \ kg$,$\Delta T_f = 0.93 \ ^\circ C$,$K_f = 1.86 \ K \ kg \ mol^{-1}$.Using the formula $\

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$C_2H_4O_2$

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(B) Given: Mass of solute $= 36 \ g$,Mass of solvent $= 1.2 \ kg$,$\Delta T_f = 0.93 \ ^\circ C$,$K_f = 1.86 \ K \ kg \ mol^{-1}$.Using the formula $\Delta T_f = K_f 	imes m$,where $m = \frac{w_2}{M_2 	imes w_1(kg)}$.Substituting the values: $0.93 = \frac{1.86 	imes 36}{M_2 	imes 1.2}$.Solving for $M_2$: $M_2 = \frac{1.86 	imes 36}{0.93 	imes 1.2} = 60 \ g \ mol^{-1}$.Empirical formula mass of $CH_2O = 12 + 2(1) + 16 = 30 \ g \ mol^{-1}$.$n = \frac{	ext{Molar mass}}{	ext{Empirical formula mass}} = \frac{60}{30} = 2$.Molecular formula $= n 	imes (CH_2O) = 2 	imes CH_2O = C_2H_4O_2$.

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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