What is the normality of 96% w/w aqueous $H_{2} S O_{4}$ solution?
(Atomic weight of $H = 1, S = 32, O = 16$ , density $= 1.84$ gm/L)

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Solution

Percentage of $W / W = 96$
$\Rightarrow 96 = \frac{Weight of H_{2} S O_{4} (W) \times 100}{Weight of H_{2} S O_{4} + Weight of water (WL)}$
$\Rightarrow \frac{96}{100} = \frac{W}{W + W_{2}}$
Solving, $W = 24 W L$
Weight of $H_{2} S O_{4} = 24$ Weight of $H_{2} O$
Normality= $\frac{Weight of H_{2} S O_{4}}{GEW of H_{2} S O_{4}} \times \frac{1000}{Volume of H_{2} S O_{4}}$
$= \frac{24 W L}{49} \times \frac{1000}{Volume of H_{2} O}$
Density= $1.84 g / L$
$\Rightarrow \frac{W_{L}}{Volume of H_{2} O} = 1.84 \Rightarrow$ Volume of $H_{2} O = \frac{W_{L}}{1.84}$
$\Rightarrow \frac{24 W_{L}}{49} \times \frac{1000}{\frac{W_{L}}{1.84}}$
$\Rightarrow \frac{24 \times 1.84 \times 1000}{49} = 901.22$

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