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Q)

$HNO_3$ used as a reagent has specific gravity of $1.42 mL^{-1}$ and contains $70\%$ by strength $HNO_3$. Calculate Normality of acid

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

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A)
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)
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