Question

The equivalent weight of `HNO_(3)` (molecular weight `=63`) in the following reaction is
`3Cu+8HNO_(3)rarr3Cu(NO_(3))_(2)+2NO+4H_(2)O`
A. `(4xx63)/(3)`
B. `(63)/(5)`
C. `(63)/(3)`
D. `(63)/(8)`

Answer 1

Correct Answer - D
In this case only `2 mol` of `NO_(3)^(ө)` undergo reduction.
`3e^(-)+NO_(3)^(ө)rarrNO(x=3)` (reduction)
`x-6= -1`
`x-2=0`
`x=5, x=2`
`6 mol` of `HNO_(3)` are not changing so `6NO_(3)^(ө)` are added in the reaction to get `3 mol` of `Cu(NO_(3))_(2)`.
`:. Ew=M+(M)/(3)=(4M)/(3)=(4xx63)/(3)`