Question

Question #586cc

Answer 1

0.4VL

Explanation:

This type of problem can be easily solved by knowing the relation between molar mass and the respective equivalent mass of each oxidant (`K_2Cr_2O_7 and KMnO_4)`.
Let me try to clear it
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In acid medium the `Cr_2O_7^(2-)` ion under goes following ion – electron reaction
`Cr_2O_7^(2-)+14H^+ +6e ->2Cr^(3+) + 7H_2O ……… (1)`

In acid medium the `MnO_4^-` ion under goes following ion – electron reaction

`MnO_4^-"" +8H^+ +5e ->Mn^(2+) +4H_2O …………….. (2)`
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Equation (1) reveals that it involves 6 electron transfer per molecule rather per formula species,So we can write
`"Equivalent mass of " K_2Cr_2O_7` = `( "Molar mass of" " " K_2Cr_2O_7)/6`
This implies that
`1(M)K_2Cr_2O_7 -=6( N) K_2Cr_2O_7`

`=>0.1(M) K_2Cr_2O_7 -=0.6(N) K_2Cr_2O_7`

So the concentration of given `K_2Cr_2O_7` solution is `S_1=0.6N` and its volume `V_1 =VL`
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Equation (2) reveals that it involves 5 electron transfer per molecule rather per formula species , So we can write

`"Equivalent mass of " KMnO_4 = ("Molar mass of """KMnO_4)/5`
This implies that
`1(M) KMnO_4-=5(N) KMnO_4`
`=>0.3(M) KMnO_4-=1.5(N) KMnO_4`

So the concentration of given `KMnO_4` solution is `S_1=1.5N "and its volume "V_2 =?`
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Here both the reactants react with same mass `0.678g" of " N_2H_4` reductant . So the product of their respective concentration in Normality and volume will be same by the law of equivalent proportion.

Hence we can write

`S_2xxV_2=S_1xxV_1=> 1.5xxV_2=0.6xxV`

`=>V_2=0.6/1.5V=2/5V=0.4VL`