# $HNO_3$ used as a reagent has specific gravity of $1.42 mL^{-1}$ and contains $70\%$ by strength $HNO_3$. Calculate Volume of water required to make $1\;N$ solution from $2\; mL$ concentrated $HNO_3?\;(\text{Normality}_{HNO_3} = 15.78)$

$(a)\;29.90 mL\qquad(b)\;29 mL\qquad(c)\;29.56 mL\qquad(d)\;30.34 mL$

Since, The molar equivalent of a solution does not change on dilution,
Molar equivalent of conc. $HNO_3$ = Molar equivalent of dilute $HNO_3$
We have, Normality of $HNO_3$ = 15.78
$\therefore 2\times15.78 = V\times1$
$\therefore$V = 31.56 mL
$\therefore$ Volume of water added = 31.56 - 2 = 29.56 mL
edited Mar 18, 2014