$100 c m^{3}$ solution is prepared by dissolving $2 g$ of NaOH in water. calculate the normality of the solution

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Solution

Solution:-
Given weight of $N a O H = 2 g$
Equivalent weight of $N a O H = 40 g$
Therefore,
Gram equivalent of $N a O H = \frac{Wt.}{Eq. wt.} = \frac{2}{40} = 0.05$
As we know that,
Normality $(N)$ $= \frac{gm eq.}{vol (in mL)} \times 1000 = \frac{0.05}{100} \times 1000 = 0.5 N$

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