A solution is made by dissolving $49 g$ of $H_{2} S O_{4}$ in $250 m L$ of water. The molarity of solution prepared is:

2 M

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1 M

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4 M

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5 M

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Solution

The correct option is A $2 M$
Molarity, $M = \frac{n o . o f m o l e s}{v o l u m e (L)}$
Volume, $V = 250 m L = 0.25 L$
No. of moles, $n =$ mass/molar mass
Mass = 49 g
Molar mass of $H_{2} S O_{4}$ = 98 g
$n = 49 / 98 = 0.5$
$M = 0.5 / 0.250 = 2 M$
Hence, option $A$ is correct.

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