Question

Calculate normality of 2.1%(w/V) $H_2{SO}_4$ solution.

• A. 2.14 N
• B. 4.28 N
• C. 0.428 N
• D. 0.214 N

Answer 1

Answer: (C)

0.428 N

2.1 % (w/v) $H_{2}S{O}_4$ means 2.1 g of$H_{2}S{O}_4$ is present in 100 ml of solution.

No. of equivalents of $H_{2}S{O}_4$ $=\frac{weight}{gramequivalentof{H}_{2}S{O}_4}=\frac{2.1}{49}moles$

Normality $=\frac{equivalentwt.}{volumeofsolution}$

$=\frac{2.1}{49}\times 1000$

= 0.428 N