Chemistry, asked by harsaajs10, 9 months ago

A mixture containing 0.05 mole of K2Cr2O7 and 0.02 mole of KMnO4 was treated with excess of KI in acidic medium. The liberated iodine required 1 L of Na2S2O3 solution for titration. What is the concentration of Na2S2O3 solution?
a. 0.4 mol/L
b. 0.2 mol/L
c. 0.25 mol/L
d. 0.3 mol/L

Answers

Answered by mohanmadhan237
4

Answer:

0.25 mol/l is answer OK

Answered by aithimouleendra
2

Answer:

Explanation:

eq. wt. of K  2

​  Cr  2   O  7

=  6

Molar mass

​eq. wt. of KMnO  4

=  5

Molar mass

​ eq. of Na  2  S  2  O  3 = eq. of I  2 liberated   = eq. of KMnO  4  + eq. of K  2 Cr2 O  7

​  or N×1=0.02×5+0.05×6  

⇒N=0.4   or      M=0.4

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