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

The equivalent weight of `H_(2)SO_(4)` in the following reaction is `N

1.	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`A. `98`B. `(98)/(6)`C. `(98)/(2)`D. `(98)/(8)`
Answer» Correct Answer - B 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)`

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

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