Question

What is the molarity of $0.2\text{N}$ ${\mathrm{H}}_2{\mathrm{SO}}_4$?

Answer 1

Step1. Formula for the calculation of molarity

$\text{Molarity}=\frac{\text{Normality}}{\text{n−factor}}$

Step2. n-factor for ${\mathrm{H}}_2{\mathrm{SO}}_4$

• n- factor of an acid or a base is the donation of number of ${\mathrm{H}}^{+}$ ions or ${\mathrm{OH}}^{-}$ ions in a reaction.
• In case of ${\mathrm{H}}_2{\mathrm{SO}}_4$, number of ${\mathrm{H}}^{+}$ ions it donates is $2$. So, the value of n-factor is $2$.

Step3. Calculation of molarity

$$\begin{gathered} \text{Molarity}=\frac{\text{0 .2}}{2} \\ =0.1\text{M} \end{gathered}$$

Hence, the molarity of $0.2\text{N}$ ${\mathrm{H}}_2{\mathrm{SO}}_4$ is $0.1\text{M}$.