Redox and Equivalent ConceptHard
Question
A sample of a metal carbonate MCO3 was neutralized by 10 ml of 0.1 N-HCl and the resulting chloride gave 0.0517 g of phosphate, M3(PO4)2. The equivalent weight of M is
Options
A.20.03
B.40.06
C.51.7
D.8.62
Solution
$n_{eq}HCl = n_{eq}M_{3}\left( PO_{4} \right)_{2}$
Or $\frac{10 \times 0.1}{1000} = \frac{0.0517}{E_{M} + 31.67} \Rightarrow E_{M} = 20.03$
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