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 $12$

$\text{Given}:$ For Calcium $(eq{)}_{Ca}$ = 2g, $(eq{)}_{H_2},released$ = 1.125L

For M metal, $(eq{)}_{H_2},released$= 1.85L

$\text{Solution}:$

$\frac{(eq{)}_{Ca}}{(eq{)}_M}=\frac{(eq{)}_{H_2},released}{(eq{)}_{H_2},released}$

$\frac{\left(\frac{2}{20}\right)}{\left(\frac{2}{x}\right)}=\frac{\frac{1.125}{(eq.vol)H_2}}{\frac{1.85}{(eq.vol)H_2}}$ [$\because$ eq wt of M=X]

$\therefore \frac{x}{20}$=$\frac{1.125}{1.85}x$ =$\frac{1.125}{1.85\times 20}\approx 12$

$\text{Hence D is the correct option}$