The mass of ascorbic acid (C_6H_8O_6) to be d | vedclass

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(C) The depression in freezing point is given by the formula: $\Delta T_f = K_f 	imes m$,where $m$ is the molality of the solution.Molality $m = \fra

Vedclass · Accepted Answer

$6.6$

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(C) The depression in freezing point is given by the formula: $\Delta T_f = K_f 	imes m$,where $m$ is the molality of the solution.Molality $m = \frac{w_2 	imes 1000}{M_2 	imes w_1}$,where $w_2$ is the mass of solute,$M_2$ is the molar mass of solute,and $w_1$ is the mass of solvent in $g$.Molar mass of ascorbic acid $(C_6H_8O_6) = (6 	imes 12) + (8 	imes 1) + (6 	imes 16) = 72 + 8 + 96 = 176 \ g \ mol^{-1}$.Given: $\Delta T_f = 1.5 \ K$,$K_f = 3.9 \ K \ kg \ mol^{-1}$,$w_1 = 100 \ g$.Substituting the values: $1.5 = \frac{3.9 	imes w_2 	imes 1000}{176 	imes 100}$.$1.5 = \frac{39 	imes w_2}{176}$.$w_2 = \frac{1.5 	imes 176}{39} = \frac{264}{39} \approx 6.77 \ g$.Rounding to the nearest provided option,the correct value is $6.6 \ g$.

The mass of ascorbic acid $(C_6H_8O_6)$ to be dissolved in $100 \ g$ of acetic acid to lower its freezing point by $1.5^{\circ}C$ in $g$ is: (Given: $K_f$ for acetic acid $= 3.9 \ K \ kg \ mol^{-1}$)

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