Analytical Chemistry Question 173
Question: 0.5 g mixture of $ K_2Cr_2O_7 $ and $ KMnO_4 $ was treated with excess of KI in acidic medium. $ I_2 $ liberated required $ 100cm^{3} $ of 0.15 N $ Na_2S_2{O_3} $ solution for titration. The percentage amount of $ K_2Cr_2O_7 $ in the mixture is
Options:
A) 85.36%
B) 14.64%
C) 58.63%
D) 26.14%
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Let the amount of the $ K_2Cr_2O_7 $ in the mixture be x g, then amount of $ KMnO_4 $ will be (0.5-x)g
$ \therefore ( \frac{x}{49}+\frac{0.5-x}{31.6} )=\frac{100\times 0.15}{1000} $ Where 49 is Eq. wt. of $ K_2Cr_2O_7 $ and 31.6 is Eq. wt. of $ KMnO_4. $ On solving, we get x = 0.073 g Percentage of $ K_2Cr_2O_7 $
$ =\frac{0.0732\times 100}{0.5}=14.64 $ %
BETA
