A certain amount of a metal whose equivalent mass is $28$ displaces $0.7 L$ of $H_{2}$ at S.T.P. from an acid, hence the mass of the element is:

1.75 g

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0.875 g

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3.50 g

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7.00 g

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Solution

The correct option is A $1.75 g$
Gram equivalent of metal = Gram equivalent of $H_{2}$

$\frac{W e i g h t o f m e t a l}{E q u i v a l e n t w t . o f m e t a l} = m o l e s o f H_{2} \times v a l e n c y f a c t o r$

$\frac{W e i g h t}{28} = \frac{g i v e n v o l u m e}{22.4} \times 2$

$\frac{W e i g h t}{28} = \frac{0.7}{22.4} \times 2$

$W e i g h t = 1.75 g m$

Hence, option $A$ is correct.

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A

V

H_{2}

E

1.80 g

833 m L

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12

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