Normality of a solution
Posted: Sat Sep 04, 2021 9:51 am
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
My answer is correct or above given answer is correct?
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
My answer is correct or above given answer is correct?