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