Normality of a solution

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Dhamnekar Winod
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Normality of a solution

Post by Dhamnekar Winod »

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.


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

My answer is correct or above given answer is correct?
Last edited by Dhamnekar Winod on Sat Sep 04, 2021 8:06 pm, edited 1 time in total.
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Re: Normality of a solution

Post by ChenBeier »

Your answer is correct.
The error is in 2/40 * 1000/250

It has time be 0,2/40 * 1000/250
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