Question

$150\, g$ of acetic acid was contaminated with $10.2 g$ ascorbic acid $\left( C _6 H _8 O _6\right)$ to lower down its freezing point by $\left( x \times 10^{-1}\right)^{\circ} C$. The value of $x$ is (Nearest integer) [Given $K _{ f }=3.9\, K\, kg \, mol ^{-1}$; Molar mass of ascorbic acid $=176\, g\, mol ^{-1}$ ]

Answer 1

$ 150 \,g \,CH _3 COOH $
$ 10.2 g \text { ascorbic acid } \Rightarrow 0.058 \text { moles } $
$ \Delta T _{ f }=\left( x \times 10^{-1}\right){ }^{\circ} C $
$ \Delta T _{ f }= k _{ f } \cdot \text { molality } $
$ =3.9 \times \frac{0.058}{150} \times 1000$
$ =1.5^{\circ} C $
$ =15 \times 10^{-1}{ }^{\circ} C $