Question

The weight of hypo $\left( Na _{2} S _{2} O _{3} \cdot 5 H _{2} O \right)$ required to make= $100\, cm ^{3}$ of $0.2\, N$ solution for use in the reaction $2 S _{2} O _{3}^{2-}+ I _{2} \longrightarrow S _{4} O _{6}^{2-}+2 I ^{-}$ will be

• A. $1.49 g$
• B. $2.98 g$
• C. $4.96 g$
• D. $8.94 g$

Answer 1

Answer: (C)

According to the given reaction,

$2 S _{2} O _{3}^{2-} \longrightarrow S _{4} O _{6}^{2-}+2 e^{-}$

$\therefore $ Eq. wt. of $Na _{2} S _{2} O _{3} \cdot 5 H _{2} O =\frac{\text { mol. wt }}{1}=\frac{248}{1}=248$

$1000 \,cm ^{3}$ of $1 N$ sol. require, $Na _{2} S _{2} O _{3} \cdot 5 H _{2} O =248 \,g$

$\therefore 100\, cm ^{3}$ of $0.2 N$ of the sol. require, $Na _{2} S _{2} O _{3} \cdot 5 H _{2} O$

$=\frac{248 \times 0.2}{1000} \times 100=4.96\, g$