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
4
Answer:
0.25 mol/l is answer OK
Answered by
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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