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Home  >>  CBSE XII  >>  Chemistry  >>  Solutions
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A sample of sulphuric acid is 98% $H_2SO_4$ by weight (density =$1.10gmL^{-1})$.Calculate normality

$\begin{array}{1 1}22N\\42N\\62N\\12N\end{array} $

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98% $H_2SO_4$ solution means that 98 g of $H_2SO_4$ are present in 100g of solution in water
Moles of $H_2SO_4=\large\frac{98}{98}$$=1$
Volume of solution =$\large\frac{\text{Mass}}{\text{Density}}$
$\Rightarrow \large\frac{100}{1.10}$$=90.9mL$
Normality =$\large\frac{\text{Gram equivalents of solute}}{\text{Volume of solution}}$$\times 1000$
Equivalent mass of $H_2SO_4=\large\frac{\text{Molar mass}}{2}$
$\Rightarrow \large\frac{98}{2}$$=49$
Basicity of $H_2SO_4=2$
Gram equivalents of $H_2SO_4=\large\frac{\text{Mass of }H_2SO_4}{\text{equivalent mass}}$
$\Rightarrow \large\frac{98}{49}=$$2$
Normality =$\large\frac{\text{Gram equivalents of }H_2SO_4}{\text{Volume of solution(mL)}}$$\times 1000$
$\Rightarrow \large\frac{2\times 1000}{90.9}$$=22N$
answered Aug 6, 2014 by sreemathi.v
 

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