Question

if 0.4g of NAOH is present in 40ml of solution what is the molarity and normality of solution​

Answer 1

$molarity = no \: of \: moles \: \div volume(in \: l) \\ m = \frac{n}{v} \\ m = \frac{given \: mass}{molecuar \: mass} \times \frac{1000}{v(in \: l)} \\ m = \frac{0.4}{23 + 16 + 1} \times \frac{1000}{40} \\ m = \frac{0.4}{40} \times \frac{100}{4} \\ m = \frac{4}{400} \times \frac{100}{4} \\ m = \frac{4}{4} \times \frac{1}{4} \\ m = \frac{1}{4} = 0.25m$

$equivalent \: mass \: of \: naoh = \frac{atomic \: weight}{valency} \\ = \frac{23}{1} = 23 \\ normality = \frac{equivalent \: weight}{molecuar \: weight} \times \frac{1000}{v(in \: \: l) } \\ = \frac{23}{40} \times \frac{1000}{40} \\ = \frac{23 \times 10}{16} \\ = \frac{230}{16} \\ = 14.3n$