The equivalent weight of `H_(2)SO_(4)` in the following reaction is
`Na_(2)Cr_(2)O_(7)+3SO_(2)+H_(2)SO_(4)rarr3Na_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+H_(2)O`

Options

A. `98`
B. `(98)/(6)`
C. `(98)/(2)`
D. `(98)/(8)`

Answer

B

Explanation

In this reaction, `6mol` of `e^(-)` are involved with `1mol` if `H_(2)SO_(4)` in this redox reaction. `H_(2)SO_(4)` acts here as acidic medium.
`6e^(-)+Cr_(2)O_(7)^(-2)rarr2Cr^(3+)(x=6)`, (oxidation)
`2x-14= -2, 2x=6`
`2x=12`
`SO_(2)rarrSO_(4)^(2-)+2e^(-)]xx3(x=6)`
`x-4=0, x-8 = -2`
`x=4, x=6`
So, `Ew of H_(2)SO_(4)=(M)/(6)=(98)/(6)`