Question

calculate the normality of an aqueous solution containing 4.9g of sulphuric acid in 500ml of solution​

Answer 1

$N=\frac{wX1000}{eXvolume in ml.}$ $,E=$ $\frac{98}{3}=32.6$

$N=\frac{4.9X1000}{32.6X500} = 0.3N$

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