MCQ
$Ca{(OH)_2} + {H_3}P{O_4} \to CaHP{O_4} + 2{H_2}O$ the equivalent weight of ${H_3}P{O_4}$ in the above reaction is
- A$21$
- B$27$
- C$38$
- ✓$49$
✓
Answer
Correct option: D.
$49$
d
(d) The equivalent weight of ${H_3}P{O_4} = \frac{{{\rm{molecular \,weight}}}}{2}$
(d) The equivalent weight of ${H_3}P{O_4} = \frac{{{\rm{molecular \,weight}}}}{2}$
mole wt of ${H_3}P{O_4}= 3 + 31 + 64 = 98$
$\frac{{98}}{2} = 49$
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