$(a)\;15.78\qquad(b)\;16.90\qquad(c)\;17\qquad(d)\;19$

Given, Strength of $HNO_3 = 70\%$

Volume of the solution = 100mL

Volume of $HNO_3$ = 70 mL

$\therefore$ Weight of $HNO_3$ in the solution = $70\times1.42g$

$\therefore$ Equivalent weight of $HNO_3$ in the solution = $\large\frac{70\times1.42}{63}$

Since N=$\large\frac{\text{Equivalent weight of $HNO_3$ in the solution}}{\text{V in litre}}$

$N_{HNO_3} = \large\frac{70\times1.42}{63\times\large\frac{100}{1000}} = 15.78$

Hence answer is (a)

Ask Question

Tag:MathPhyChemBioOther

Take Test

...