Correct Answer – A
`undersetulbar(Cr(2)O_(7)^(2-)+14H^(+)2Sn^(2+) rarr 3Sn^(4+)+2Cr^(3+)+7H_(2)O)(Cr_(2)O_(7)^(2-)+14H^(+)+underset((Sn^(2+) rarr Sn^(4+)+2e^(-))xx3)(6e^(-)to2Cr^(3+)+7H_(2)O)`
It is clear form this equation that `3` moles of `Sn^(2+)` reduce one mole of `Cr_(2)O_(7)^(2-)`, hence `1` mol. Of `Sn^(2+)` will reduce `(1)/(3)`moles of `Cr_(2)O_(7)^(2-)`.

Correct Answer – D
`{:(Cr_(2)O_(7)^(2-),rarr,Cr^(3+),Sn^(2+),rarr,Sn^(4+)),(Cr(+6),,Cr(+3),Sn(+2),,Sn(+4)):}`
One formula unit of dichromate ion contains two Cr atoms. Thus, total change in oxidation number is of 6 units `(+12` to `+6)` . Thus, 1mole of `Cr_(2)O_(7)^(2-)` contains 6 equivalents. For tin, the oxidation number change by 2 units. Thus, 1 mole of `Sn^(2+)` ion contains 2 equivalents. According to law of equivalents, two equivalentsof `Sn^(2+)` ion will only reduce 2 equivalents of `Cr_(2)O_(7)^(2-)`.
6 equivalents of `Cr_(2)O_(7)^(2-)` come form 1 mole
1 equivalent will come form `(1)/(6)` mole
2 equivalents will come form `(2)/(6)` mole i.. `1//3` mole.

Correct Answer – A
`(6e^(-)+Cr_(2)O_(7)^(2-)rarr2Cr^(3+))`
`Sn^(2+)rarrSn^(4+)+2e^(-)`
Equivalent of `Cr_(2)O_(7)^(2-)= “Equivalent of” Sn^(2+)`
`(n=6)` , `(n=2)`
`implies1 Eq= 1Eq`
`(1)/(6) mol =(1)/(2) mol`
`(1)/(3) “mol of” Cr_(2)O_(7)^(2-)= 1 “mol of” Sn^(2+)`.

Correct Answer – A
`Cr_(2)O_(7)^(2-)+3Sn^(2+)+14H^(+)rarr 2 Cr^(3+)+3Sn^(4+)+7H_(2)O`
1 mole of `Sn^(2+)` will reduce `1/3` moles of `K_(2)Cr_(2)O_(7)`.

Correct Answer – C
`undersetulbar(Cr(2)O_(7)^(2-)+14H^(+)2Sn^(2+) rarr 3Sn^(4+)+2Cr^(3+)+7H_(2)O)(Cr_(2)O_(7)^(2-)+14H^(+)+underset((Sn^(2+) rarr Sn^(4+)+2e^(-))xx3)(6e^(-)to2Cr^(3+)+7H_(2)O)`
It is clear form this equation that `3` moles of `Sn^(2+)` reduce one mole of `Cr_(2)O_(7)^(2-)`, hence `1` mol. Of `Sn^(2+)` will reduce `(1)/(3)`moles of `Cr_(2)O_(7)^(2-)`.