Question

The normality of a mixture obtained by mixing $0.62$ g of ${Na}_{2}C{O}_3\cdot H_{2}O$ to $100$ mL of $0.1N{H}_{2}S{O}_4$ is (write it in multiples of $10$ (only single digit).

Answer 1

The molar mass of $N{a}_{2}C{O}_3.H_{2}O$ is 124 g/mol.

The number of g equivalents present in 0.62 g $N{a}_{2}C{O}_3.H_{2}O$ are $\frac{2\times 0.62}{124}=0.010geq$

The number of g eq present in 100 ml of 0.1 N $H_{2}S{O}_4$ are $\frac{100ml}{1000ml/L}\times 0.1=0.01geq$

$N{a}_{2}CC{O}_3+H_{2}S{O}_4\rightleftharpoons N{a}_{2}S{O}_4+H_{2}C{O}_3$

Thus, 0.010 g eq of $N{a}_{2}C{O}_3$ reacts with 0.01 g eq of $H_{2}S{O}_4$ to form 0.01 g eq of $N{a}_{2}S{O}_4$ and 0.01 g eq of $H_{2}C{O}_3$.

Hence, the normality of the mixture is $\frac{0.01geq}{\frac{100ml}{1000ml/L}}=0.1N$.

Hence, the normality of the mixture will be 0.1 N.