Explanation

(b)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ N_{1}V_{1} = N_{2}V_{2}$
Given, $N_{1}$ = 10 N $HNO_{3},$ $N_{2}$ = 0.1 N $HNO_{3}$
$V_{1}$= 10 mL, $V_{2}$ =?
$\therefore\ \ 10\ \times 10 = 0.1 \times \ V_{2}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ V_{2}\ \ = \frac{10\ \times \ 10}{0.1}$
= 1000mL
$\therefore$ 10 mL water is already there in solution.
$\therefore$ Water to be added =1000-10
=990mL