Equivalent weight of a divalent metal is $24$ . The volume of hydrogen liberated at STP by $12 g$ of the same metal when added to excess of an acid solution is:

2.8 l i t e r s

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5.6 l i t e r s

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11.2 l i t e r s

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22.4 l i t e r s

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Solution

The correct option is B $5.6 l i t e r s$
Equivalent weight $= \frac{m o l e c u l a r w e i g h t}{v a l e n c y}$

$24 = \frac{M_{w}}{2}$

$M_{w} = 48 =$ Molecular weight of divalent metal $X$

Moles of $X = \frac{12}{48} = 0.25$

$X + 2 H C l \to X C l_{2} + H_{2}$

$0.25$ $0.50$ $-$ $-$

$0.25$ $0.25$

Moles of $H_{2}$ produced $= 0.25 m o l e s$

Volume of $0.25 m o l e s H_{2} = 0.25 \times 22.4 = 5.6 l i t r e s$

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Equivalent weight of a divalent metal is 24. The volume of hydrogen liberated at STP by 12 g of the same metal when added to excess of an acid solution is:

The correct option is B 5.6 litersEquivalent weight =molecular weightvalency 24=Mw2 Mw=48= Molecular weight of divalent metal X Moles of X=1248=0.25 X+2HCl→XCl2+H2 0.25 0.50 − − 0.25 0.25 Moles of H2 produced =0.25 moles Volume of 0.25 moles H2=0.25×22.4=5.6 litres

Equivalent weight of a divalent metal is $24$ . The volume of hydrogen liberated at STP by $12 g$ of the same metal when added to excess of an acid solution is: