Answer
Please see the work below.
Work Step by Step
We know that:
$molar\space mass\space of CHCl_3=12.01+1.008+3(35.45)=119.4\frac{g}{mol}$
Now, $\%C=\frac{12.01}{119.4}\times 100\%=10.06\%$
$\%H=\frac{1.008}{119.4}\times 100\%=0.8442\%$
$\%Cl=\frac{3(35.45)}{119.4}\times 100\%=89.07\%$
