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Q.

$Ca (OH)_{2} + H_{3} {PO}_{4} \to {CaHPO}_{4} + 2 H_{2} O$ the
weight of $H_{3} P 0_{4_{}}$ in the above reaction is

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a

27

b

38

c

21

d

49

answer is D.

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Detailed Solution

The equivalent weight of $H_{3} {PO}_{4} = \frac{molecular weight}{2}$

$∵$ mole wt of $H_{3} {PO}_{4} = 3 + 31 + 64 = 98$

$∴ \frac{98}{2} = 49$

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