The depression in freezing point observed for a formic acid solution of concentration 0 . 5   mL  …

$\left[\mathrm{HCOOH}\right]=0.5\mathrm{ml} {\mathrm{L}}^{-1}$

$\Rightarrow \left(0.5\mathrm{ml}\times 1.05 \mathrm{g} {\mathrm{ml}}^{-1}\right) \mathrm{HCOOH}$ in $1 \mathrm{L}$

$\Rightarrow 0.525 \mathrm{g} \mathrm{HCOOH}$ in $1 \mathrm{L}$

Weight of water will be $= 1 \mathrm{kg} \mathrm{since} \mathrm{density} \mathrm{of} \mathrm{water} \mathrm{is} 1 \mathrm{kg}/\mathrm{L}$

$\mathrm{m}=\frac{\left({\displaystyle \frac{0.525}{46}}\right)}{1 \mathrm{kg}} \mathrm{mol}$ [Assuming dilute solution]

$$\begin{gathered} \therefore {\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{iK}}_{\mathrm{f}}\mathrm{m}\Rightarrow \\ \mathrm{i}=\frac{{\mathrm{\Delta T}}_{\mathrm{f}}}{{\mathrm{k}}_{\mathrm{f}}\mathrm{m}}=\frac{0.0405\times 46}{1.86\times 0.525}=1.9 \end{gathered}$$