A mixture containing 0.05 mole of $K_{2} {Cr}_{2} O_{7}$ , and 0.02 mole of KMnO₄ was treated with excess of Kl in acidic medium. The liberated iodine required 1.0 L of Na₂S₂O₃ solution for titration. Concentration of Na₂S₂O₃ solution was:

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0.40 mol L^-1

0.20 mol L^-1

0.25 mol L^-1

0.30 mol L^-1

answer is A.

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Detailed Solution

eq. wt. of $K_{2} {Cr}_{2} O_{7} = \frac{molar mass}{6}$
eq. wt. of ${KMnO}_{4} = \frac{molar mass}{5}$
eq. of ${Na}_{2} S_{2} O_{3}$ = eq. of $I_{2}$ libeIated
= eq. of KMnO₄ + eq. of K₂Cr₂O₇
or N x 1= 0.02 x 5 + 0.05 x 6
= N = 0.4 or M=0.4

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