Question

The amount of ${Na}_2{S}_2{O}_3\cdot 5H_{2}O$ required to completely reduce $100mL$ of $0.25N$ iodine solution, is:

• A. $6.20g$
• B. $9.30g$
• C. $3.10g$
• D. $7.75g$

Answer 1

Answer: (A)

$6.20g$

$N{a}_2{S}_2{O}_3+I_2\to N{a}_2{S}_2{O}_6+NaI$

Equivalents of $N{a}_2{S}_2{O}_3$ = equivalents of $NaOH$

$1\times moles=0.25\times 100\times {10}^{-3}$

Moles = 0.025

weight of $N{a}_2{S}_2{O}_3.5H_{2}O=0.025\times 248=6.2gm$