Question 586cc

May 14, 2016

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 $C {r}_{2} {O}_{7}^{2 -}$ ion under goes following ion – electron reaction
Cr_2O_7^(2-)+14H^+ +6e ->2Cr^(3+) + 7H_2O ……… (1)

In acid medium the $M n {O}_{4}^{-}$ 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
$\text{Equivalent mass of } {K}_{2} C {r}_{2} {O}_{7}$ = $\frac{\text{Molar mass of" " } {K}_{2} C {r}_{2} {O}_{7}}{6}$
This implies that
$1 \left(M\right) {K}_{2} C {r}_{2} {O}_{7} \equiv 6 \left(N\right) {K}_{2} C {r}_{2} {O}_{7}$

$\implies 0.1 \left(M\right) {K}_{2} C {r}_{2} {O}_{7} \equiv 0.6 \left(N\right) {K}_{2} C {r}_{2} {O}_{7}$

So the concentration of given ${K}_{2} C {r}_{2} {O}_{7}$ solution is ${S}_{1} = 0.6 N$ and its volume ${V}_{1} = V L$
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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
$1 \left(M\right) K M n {O}_{4} \equiv 5 \left(N\right) K M n {O}_{4}$
$\implies 0.3 \left(M\right) K M n {O}_{4} \equiv 1.5 \left(N\right) K M n {O}_{4}$

So the concentration of given $K M n {O}_{4}$ solution is S_1=1.5N "and its volume "V_2 =? #
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Here both the reactants react with same mass $0.678 g \text{ of } {N}_{2} {H}_{4}$ 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

${S}_{2} \times {V}_{2} = {S}_{1} \times {V}_{1} \implies 1.5 \times {V}_{2} = 0.6 \times V$

$\implies {V}_{2} = \frac{0.6}{1.5} V = \frac{2}{5} V = 0.4 V L$