($k _{ f }=1.86\,K\,kg\,mol ^{-1}$ )
$\Rightarrow\left(0.5 ml \times 1.05\,g\,ml ^{-1}\right) HCOOH$ in $1 L$
$\Rightarrow 0.525\,g\,HCOOH$ in $1\,L$
$m =\frac{(0.525 / 46)}{1\,kg } mol$ [Assuming dilute solution]
$\therefore \Delta T _{ f }= iK _{ f } m \Rightarrow i =\frac{\Delta T _{ f }}{ k _{ f } m }=\frac{0.0405 \times 46}{1.86 \times 0.525}=1.9$
(ઉપયોગ $R$ $=0.083\,L\,bar\,K ^{-1}\,mol ^{-1}$ )