## Normality of a solution

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Dhamnekar Winod
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### Normality of a solution

My answer to the following question is 0.02 N But the answer given is 0.2N. Where I am wrong?

Question :Calculate the normality of NaOH solution formed by dissolving 0.2 gm NaOH to make 250 ml solution.

Solution:

Normality (N) = $$\frac{number of Gram Equivalent of solute}{Volume of Solution in litre}$$
Number of Gram Eq. of the Solute = $$\frac{weight}{Equivalent weight}$$
Now, Equivalent weight= Molar Mass n =23+16+11=40

So, N =$$\frac{ No.of gram eq.mass}{Vol (liter)}$$
$$= \frac{Weight}{Equivalent weight × 1000/V(in ml)}$$
= $$\frac{2}{40} \times \frac{1000}{250}$$
= 0.2 N

Last edited by Dhamnekar Winod on Sat Sep 04, 2021 8:06 pm, edited 1 time in total.
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