Question

The Equivalent weight of $H_{3}P{O}_4$ in the reaction is :
$Ca(OH{)}_2+H_{3}P{O}_4\to CaHP{O}_4+2H_{2}O$ $(Ca=40,P=31,O=16)$

• A. 49
• B. 98
• C. 32.66
• D. 147

Answer 1

The correct option is A 49

$Ca(OH{)}_2+H_{3}P{O}_4\to CaHP{O}_4+2H_{2}O$

Since two H-atoms are replaced in the reaction.
$\therefore Ew=\frac{Mw}{n}=\frac{98}{2}=49$