0.49 grams of Sulphuric acid is dissolved in 500 ml solution. Molarity of the solution is

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0.05 M

0.01 M

0.02 M

0.025 M

answer is A.

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Detailed Solution

$M = n \times \frac{1}{V (l)}; M = n \times \frac{1000}{V (ml)}; n = \frac{w_{solute}}{GMW of solute}; M = \frac{0.49}{98} \times \frac{1000}{500} = 0.01 M;$

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