Explanation

(a)
$N_{1}V_{1} = N_{2}V_{2}$
0.164 M NaOH $\cong$ 0.164 N NOH
$N_{1}$ = ?, $V_{1}\$= 25 mL, $N_{2}$ = 0.164, $V_{2}$ =32.63 mL
${\ \ \ \ \ \ N}_{1}V_{1} = N_{2}V_{2}$
or ${\ \ \ \ N}_{1} =$ $\frac{N_{2}V_{2}\ }{V_{1}}$
= $\frac{0.164\ \times \ 32.63}{25}$
= 0.214 N $H_{2}{SO}_{4}$
0.214 N $H_{2}{SO}_{4} \cong \frac{0.214}{2}\ M\ \ H_{2}{SO}_{4}$ $(\because\ Basicity\ of\ \ H_{2}{SO}_{4} = 2)$
$\ \ \ \ \ \ \cong$0.107 M $H_{2}{SO}_{4}$