Explanation

(b)
$N_{1}V_{1} = N_{2}V_{2}$
Amount of water to be added
= total volume – volume of NaOH
Given, normality of NaOH =$N_{1}$ =0.1 N
Volume of NaOH =$V_{1}$=?
Normality of HCl($N_{2}$)= 0.2 N
Volume Of HCl = $V_{2}$ = 50mL
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ N_{1}V_{1} = N_{2}V_{2}$
$0.1 \times \ V_{1} = 0.2 \times \ 50\ \$
${\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V}_{1} = \frac{0.2 \times \ 50}{0.1} =$ 100mL
V of NaOH = 40 mL
Amount of water to be added =100 - 40=60 mL