$1.2$ g of $M g$ is treated with $100$ mL of $1$ M $H_{2} S O_{4}$ . Molar concentration of the $H_{2} S O_{4}$ solution after complete reaction is :

0.5

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0.005

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0.10

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0.20

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Solution

The correct option is A $0.5$ M
$M g + H_{2} S O_{4} \to M g S O_{4} + H_{2}$
$1.2$ g $M g$ reacts with $0.05$ mol $H_{2} S O_{4}$ .
$H_{2} S O_{4}$ taken $= 100$ mL of $1$ M $H_{2} S O_{4}$ $= 0.1$ mol.
$H_{2} S O_{4}$ left $= 0.1 - 0.05 = 0.05$ mol in $100$ mL solution.
[ $H_{2} S O_{4}$ ] = $0.05$ $\times$ $10 = 0.5$ M.

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1.2 g of Mg is treated with 100 mL of 1 M H2SO4. Molar concentration of the H2SO4 solution after complete reaction is :

The correct option is A 0.5 MMg+H2SO4→MgSO4+H21.2 g Mg reacts with 0.05 mol H2SO4.H2SO4 taken =100 mL of 1 M H2SO4 =0.1 mol. H2SO4 left =0.1−0.05=0.05 mol in 100 mL solution. [H2SO4] = 0.05 × 10=0.5 M.

$1.2$ g of $M g$ is treated with $100$ mL of $1$ M $H_{2} S O_{4}$ . Molar concentration of the $H_{2} S O_{4}$ solution after complete reaction is :