$4.9 g$ of $H_{2} S O_{4}$ is present in $100 m L$ of the solution. What is the molarity of the solution? Calculate its normality also.

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Solution

Strength $(g / L)$ of $H_{2} S O_{4} = \frac{4.9}{100} \times 1000 = 49$
$M o l a r i t y = \frac{S t r e n g t h}{M o l . m a s s} = \frac{49}{48} = 0.5 M$
$N o r m a l i t y = n \times M o l a r i t y$
$n = \frac{M o l . m a s s}{E q . m a s s} = \frac{98}{49} = 2$
$N o r m a l i t y = 2 \times 0.5 = 1.0$
Hence, the $H_{2} S O_{4}$ solution is $1 N$ .

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