Question

Calculate the mass of ascorbic acid (Vitamin C. ${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{6}}$) to be dissolved in 75 g of acetic acid to lower its melting point by $\text{1}{\text{.5}}^{\circ }\text{C}\text{.}{\text{K}}_f=3.9\text{K}\text{kg}\text{mo}{\text{l}}^{-1}$.

Answer 1

Mass of acetic acid, w1= 75 g Molar mass of ascorbic acid ($C_6{H}_8{O}_6$),

$M2=6\times 12+8\times 1+6\times 16$

$=176gmo{l}^{-1}$

Lowering of melting point, $\Delta T_f=1.5K$

We know that:

$\Delta T_b=\frac{K_b\times 1000\times w_2}{M_2\times w_1}$

$w_2=\frac{\Delta T_b\times M_2\times w_1}{K_b\times 1000}$

$w_2=\frac{1.5\times 176\times 75}{3.9\times 1000}$

$w_2=5.08g$

Hence, 5.08 g of ascorbic acid is needed to be dissolved.