No. of equivalents of $NaOH =50 \times 0.1=5$
No. of equivalents of $H _{2} SO _{4}$ left $=20-5=15$
$150 \times x =15$
$x =\frac{1}{10}=0.1 N =1 \times 10^{-1}\,N$
(for $AgCl$ માટે $K_{sp}$ $= 1.8 \times 10^{-10},$ for $PbCl_2$ માટે $ K_{sp}$ $= 1.7 \times 10^{-5}$)