A) \[\frac{M}{3}\]
B) \[\frac{M}{6}\]
C) \[\frac{M}{7}\]
D) \[\frac{M}{8}\]
Correct Answer: D
Solution :
[d] \[\underset{\underset{-3}{\mathop{\uparrow }}\,}{\mathop{C{{u}_{3}}P}}\,\to \underset{\underset{+5}{\mathop{\uparrow }}\,}{\mathop{{{H}_{3}}P{{O}_{4}}}}\,\] Change in oxidation number =8 units of phosphorus \[\therefore \]Equivalent weight of \[{{H}_{3}}P{{O}_{4}}=\frac{M}{8}\]
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