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The `Ew` of `H_(3) PO_(4)` in reaction is `Ca(OH)_(2) + H_(3) PO_(4) rarr CaHPO_(4) + 2H_(2)O` `(Ca = 40, P = 31, O = 16)`

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The `Ew` of `H_(3) PO_(4)` in reaction is
`Ca(OH)_(2) + H_(3) PO_(4) rarr CaHPO_(4) + 2H_(2)O`
`(Ca = 40, P = 31, O = 16)`
A. 49
B. 98
C. 32.66
D. 147
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Correct Answer - A
Since two `H`- atoms are replaced
`:. Ew = (Mw)/(n) = (98)/(2) = 49`
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