Question

Complete reaction of 2.0 g of calcium (at. wt. = 40) with excess HCL produces 1.125 L of $H_2$ gas. Complete reaction of the same quantity of another metal "M" with excess HCL produces 1.85 L of $H_2$ gas under identical conditions. The equivalent weight of "M" is closest to

• A. 23
• B. 9
• C. 7
• D. 12

Answer 1

The correct option is D

12

1.125L of H${}_2$ produced by 0.1 equivalent of metal

1.85L a of H${}_2$ will be produced by =$\frac{0.1\times 1.85}{1.125}$ equivalents

$\therefore$No of gram equivalent of metal

$=\frac{2}{\text{Equivalent weight}}=\frac{2}{x}$

$\therefore \frac{0.1}{1.125}\times 1.85=\frac{2}{x}$

X=12.16