Question

1 g-atom of a divalent metal reacts completely with a dilute acid. The volume of $H_2$ liberated at S.T.P. would be?

• A. ${22.4dm}^3$
• B. ${11.2dm}^3$
• C. ${44.8dm}^3$
• D. ${5.6dm}^3$

Answer 1

Answer: (A)

${22.4dm}^3$

Gram-atom is also called as atomic weight in grams. The reaction of divalent metal with a dilute acid:

$M+$ $2H^{+}=$$H_2$ $+$ $M^{2+}$

So 1 g-atom i,e. 1 mole of divalent metal liberates 1 mole of $H_2$. The volume of $H_2$ liberated at S.T.P. would be $22.4L$ i.e.${22.4dm}^3$.

So the correct answer is A.