Question

$100{\mathrm{cm}}^3$solution is prepared by dissolving $\text{'}2\mathrm{g}\text{'}$of $\mathrm{NaOH}$ in the water. Calculate the normality of the solution.

Answer 1

Formula to calculate Normality:

$\mathrm{Normality}\left(\mathrm{N}\right)=\frac{\mathrm{Given}\mathrm{weight}\mathrm{of}\mathrm{substance}}{\mathrm{Equivalent}\mathrm{weight}\mathrm{of}\mathrm{substance}}\times \frac{1000}{\mathrm{Volume}(\mathrm{in}\mathrm{mL})}$

Given the weight of $\mathrm{NaOH}=2\mathrm{g}$

The molecular weight of $\mathrm{NaOH}=40\mathrm{g}$
The equivalent weight of $\mathrm{NaOH}=\frac{\mathrm{Molecular}\mathrm{weight}}{\mathrm{Acidity}\mathrm{of}\mathrm{Base}}=\frac{40}{1}=40$

The Given volume of solution$=100{\mathrm{cm}}^3=100\mathrm{mL}$

$\mathrm{Normality}\left(\mathrm{N}\right)=\frac{2}{40}\times \frac{1000}{100}=0.5\mathrm{N}$

Conclusion statement: The normality of the solution of the given solution is $0.5\mathrm{N}$