The equivalent weight of ${H_{3} P O}_{4}$ in the following reaction is:

${H_{3} P O}_{4} + C a {(O H)}_{2} ⟶ {C a H P O}_{4} + 2 H_{2} O$

98

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49

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32.66

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40

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Solution

The correct option is B $49$
$E q u i v a l e n t w e i g h t = \frac{M o l e c u l a r w e i g h t}{No. of replaceable Hydrogen atoms .}$

$= \frac{98}{2}$

$= 49$

Option B is correct.

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Similar questions

H_{3} P O_{4}

C a (O H)_{2} + H_{3} P O_{4} \to C a H P O_{4} + 2 H_{2} O

(C a = 40, P = 31, O = 16)

The equivalent weight of H3PO4 in following reaction

H3PO4 + Ca(OH)2 ---> CaHPO4 +2 H2O

C a (O H)_{2} + H_{3} P O_{4} \to C a H P O_{4} + 2 H_{2} O

H_{3} P O_{4}

H_{3} P O_{4}

C a (O H)_{2} + H_{3} P O_{4} \to C a H P O_{4} + 2 H_{2} O

C a (O H)_{2} + H_{3} P O_{4} \to C a H P O_{4} + 2 H_{2} O

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