Question

To a $25 \,ml \,H _{2} O _{2}$ solution, excess of acidified solution of KI was added. The iodine liberated required $20\, ml$ of $0.3 \,N \,Na _{2} S _{2} O _{3}$ solution.

• A. 1.344 g/L
• B. 3.244 g/L
• C. 5.4 g/L
• D. 4.08 g/L

Answer 1

Answer: (D)

4.08 g/L

Answer 2

$2 K I + H _{2} SO _{4}+ H _{2} O _{2} \longrightarrow K _{2} SO _{4}+2 H _{2} O + I _{2}$ $2 Na _{2} S _{2} O _{3}+ I _{2} \longrightarrow Na _{2} S _{4} O _{6}+2 Nal$ milli e of $H _{2} O _{2}$ in $50\, ml =$ milli e of $I _{2}$ $=$ milli e of $Na _{2} S _{2} O _{3}$ milli e of $H _{2} O _{2}$ in $25\, ml =20 \times 0.3=6$ milli e of $H _{2} O _{2}$ in $1000\, m 1=\frac{6}{25} \times 1000=240$ Equivalent $=\frac{240}{1000}$ Gram per litre of $H _{2} O _{2}=\frac{240}{1000} \times 17=4.08 \,g /\, L$ (Equivalent weight of $\left.H _{2} O _{2}=\frac{34}{2}=17\right)$.