The solution contains $1.25 \times 10^{- 3}$ kg of $H_{2} S O_{4}$ . The volume of $500 c m^{- 3}$ . Find the normality of solution. Molecular weight of $H_{2} S O_{4}$ is $98 g / m o l e$ .

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Solution

$N o r m a l i t y = \frac{m a s s}{e q u i v a l e n t m a s s} \times \frac{1000}{V (i n m l)}$
Equivalent mass of $H_{2} S O_{4} = \frac{98}{2} = 49 g$
Normality of $H_{2} S O_{4}$ solution = $\frac{1.25 g}{49 g} \times \frac{1000}{500} = \frac{2.5}{49} = 0.05 N$

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The solution contains 1.25×10−3 kg of H2SO4. The volume of 500cm−3. Find the normality of solution. Molecular weight of H2SO4 is 98 g/mole.

Normality=massequivalentmass×1000V(inml)Equivalent mass of H2SO4=982=49gNormality of H2SO4 solution = 1.25g49g×1000500=2.549=0.05N

The solution contains $1.25 \times 10^{- 3}$ kg of $H_{2} S O_{4}$ . The volume of $500 c m^{- 3}$ . Find the normality of solution. Molecular weight of $H_{2} S O_{4}$ is $98 g / m o l e$ .

$N o r m a l i t y = \frac{m a s s}{e q u i v a l e n t m a s s} \times \frac{1000}{V (i n m l)}$
Equivalent mass of $H_{2} S O_{4} = \frac{98}{2} = 49 g$
Normality of $H_{2} S O_{4}$ solution = $\frac{1.25 g}{49 g} \times \frac{1000}{500} = \frac{2.5}{49} = 0.05 N$