Analytical Chemistry Question 269
Question: To a 25 ml of $ H_2O_2 $ solution, excess of acidified solution of KI was mixed. The liberated I2 require 20ml of 0.3M hypo solution for neutralization. The volume strength of $ H_2O_2 $ will be
[MP PET 2003]
Options:
A) 1.34 ml
B) 1.44 ml
C) 1.60 ml
D) 2.42 ml
Show Answer
Answer:
Correct Answer: A
Solution:
20 ml of $ 0.3N $
$ Na_2S_2O_3 $
$ =20ml $ of $ 0.3NI_2 $ Solution $ =20ml $ of $ 0.3NH_2O_2 $ solution $ \equiv 25ml $ of $ 0.08N H_2O_2 $ solution Mass of $ H_2O_2 $
$ 100ml $ solution = $ \frac{0.08\times 17\times 100}{1000} $
$ =0.136gm $ % = 0.136 $ 68gmH_2O_2 $ evolve oxygen at NTP $ =22400ml $
$ 0.00136gmH_2O_2 $ evolve oxygen at NTP $ =\frac{22400}{68}\times 0.00136=0.448 $ For $ 0.1N $ , the solution is of 0.448 volume.
$ \therefore $ $ 3N $ , volume $ =0.448\times 3=1.344\text{ mL} $ .
BETA
