100mL of $H_{2} O_{2}$ solution having volume strength 11.35 V is mixed with 500 mL of 0.5 M KI solution to liberate $I_{2}$ gas such that equilibria gets established. All the $I_{2}$ gas liberated is dissolved to form 500 mL solution. 200 ml of the solution required 50 mL of $\frac{2}{3} M$ hypo solution. Calculate volume strength of remaining $H_{2} O_{2}$ mixture. Round off your answer.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 1.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\underset{n f = 2}{H_{2} O_{2}} + \underset{n f = 1}{K I} \to \underset{n f = 2}{I_{2}}$
$\underset{2 \times m m I_{2}}{\underset{n f = 2}{I_{2}}} + \underset{= 1 \times m m N a_{2} S_{2} O_{3}}{\underset{n f = 1}{N a_{2} S_{2} O_{3}}} \to N a_{2} S_{4} O_{6}$
$2 \times m m I_{2} = 1 \times m m N a_{2} S_{2} O_{3}$
$m m I_{2} = 1 \times \frac{50}{2} \times \frac{2}{3} = \frac{50}{3} m m$

In 500 ml of solution mm of $I_{2} = \frac{50}{3} \times \frac{5}{2} = \frac{125}{3} m m$
mm of $H_{2} O_{2} \times 2 = 2 \times m m I_{2}$
mm of $H_{2} O_{2} = \frac{125}{3}$ reacted
mm of $H_{2} O_{2}$ left $= 100 - \frac{125}{3} = \frac{175}{3} m m$
$M = \frac{175}{3 \times 600} = 0.097$
$V S = 0.097 \times 11.35 = 1.1$