Question

What is the strength in gram per litre of a solution of $H_{2}S{O}_4$, $12$ mL of which is neutralized by $15$ mL of $\frac{N}{10}$ $NaOH$ solution?

• A. $6.125$
• B. $8.215$
• C. $6.625$
• D. $8.125$

Answer 1

Answer: (A)

$6.125$

milliequivalents of $H_{2}S{O}_4=$ milli equivalents of NaOH
$N_1{V}_1=N_2{V}_2$
$N_1\times 12=\frac{1}{10}\times 15$
$\therefore N_1=0.125$
$Strength=N\times Eq=0.125\times 49=6.125g{L}^{-1}$
[Mw of $H_{2}S{O}_4=1\times 2+32+16\times 4=98g$]
$[Ew=\frac{Mw}{2}=\frac{98}{2}=49g]$