Question #586cc

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#