Question

How many milliliters of $0.1N{H}_{2}S{O}_4$ solution will be required for complete reaction with a solution containing $0.125g$ of pure $N{a}_{2}C{O}_3$?

• A. $23.6mL$
• B. $25.6mL$
• C. $26.3mL$
• D. $32.6mL$

Answer 1

The correct option is A $23.6mL$

No. of moles of $N{a}_{2}C{O}_3=\frac{Weight}{Molecularweight}=\frac{0.125}{106}=0.001179$ $mol$

No. of equivalents of $N{a}_{2}C{O}_3=$ (No. of moles)(Valence factor) $=0.001179\times 2=0.002358$

It gives number of equivalents required for complete reaction.

For, $H_{2}S{O}_4$, No. of equivalents $=$ Normality $\times$ Volume

$\Rightarrow 0002358=0.1\times V$

$\Rightarrow V=23.58$ $mL$

$\Rightarrow V=23.6$ $mL$