Question

# A mixture containing $$0.05\ mole$$ of $$K_{2}Cr_{2}O_{7}$$ and $$0.02\ mole$$ of $$KMnO_{4}$$ was treated with excess of $$KI$$ in acidic medium. The liberated iodine required $$1.0\ L$$ of $$Na_{2}S_{2}O_{3}$$ solution for titration. Concentration of $$Na_{2}S_{2}O_{3}$$ solution was ?

A
0.40 mol L1
B
0.20 mol L1
C
0.25 mol L1
D
0.30 mol L1

Solution

## The correct option is A $$0.40\ mol\ L^{-1}$$eq. wt. of $$K_2Cr_2O_7 = \dfrac{{\text{Molar mass}}}{6}$$ eq. wt. of $$KMnO_4 = \dfrac{{\text{Molar mass}}}{5}$$ eq. of $$Na_2S_2O_3$$ = eq. of $$I_2$$ liberated     = eq. of $$KMnO_4$$ + eq. of $$K_2Cr_2O_7$$ or $$N \times 1 = 0.02 \times 5 + 0.05 \times 6$$ $$\Rightarrow \, N = 0.4$$   or      $$M = 0.4$$ Chemistry

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