Chemistry, asked by harpreet1252, 6 months ago

Caluclate the normality of 500ml of solution containg 5.3g of sodium bi carbonate

Answers

Answered by netalsheth

1

Answer

0.216N

Explanation:

N=w×1000/E×V

w=weight of solute

E=equivalent mass

V=volume of solution in ml

so, N=5.3×1000/84×500 (Eq .mass of NaHCo3=84)

Ans is 0.216N

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