150   g of acetic acid was contaminated with 10 . 2   g ascorbic acid C 6 H 8 O 6 to lower down…

Given $150 \mathrm{g} \mathrm{of} \mathrm{solvent} {\mathrm{CH}}_3\mathrm{COOH}$

And as solute $10.2 \mathrm{g}$ ascorbic acid $=\frac{10.2}{176} = 0.058$ moles

${\mathrm{\Delta T}}_{\mathrm{f}}=\left(\mathrm{x}\times {10}^{-1}\right)°\mathrm{C}$

${\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{k}}_{\mathrm{f}}·$ molality

$=3.9\times \frac{0.058}{150}\times 1000$

$=1.5 °\mathrm{C}$

$=15\times {10}^{-1} °\mathrm{C}$

So, value of x=15.