1.

What is the equivalent weight of `NH_(3)` in the given reaction? `3CuO+2NH_(3) rarr 3Cu+N_(2)+3H_(2)O`A. `17`B. `(17)/(4)`C. `(17)/(2)`D. `(17)/(3)`

Answer» Correct Answer - D
`2N^(3-) rarr (N_(2))^(0)+6e^(-)`
`2` mole of `NH_(3)=1` mole `N_(2)`
Thus equivalents
`2xxn=1xx6`
`n=6//2=3`
`:.` Eq. wt. `= (M)/(3)=(17)/(3)`


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