The equivalent weight of $N a_{2} S_{2} O_{3}$ (Mol. wt . = M ) in the reaction,

$2 N a_{2} S_{2} O_{3} + I_{2} \to N a_{2} S_{4} O_{6} + 2 N a I$ is :

M/4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

M/3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

M/2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D M
$\begin{matrix} + 2 {2 N a}_{2} S_{2} O_{3} \end{matrix} + I_{2} ⟶ \begin{matrix} 5 / 2 {N a}_{2} S_{4} O_{6} \end{matrix} + 2 N a I 2 + 2 x - 6 = 02 + 4 x - 12 = 0 2 x - 4 = 04 x - 10 = 0 2 x = 44 x = 10 x = + 2 x = \frac{10}{4} = \frac{5}{2}$

Change in O.S.= $\frac{5}{2} - \frac{2}{1} = \frac{5 - 4}{2} = \frac{1}{2}$

$n - f a c t o r o f {N a}_{2} S_{2} O_{3} = \frac{1}{2} \times 2 = 1$

$E q u i v a l e n t w e i g h t = \frac{M o l . w t .}{n - f a c t o r} = \frac{M}{1}$

Hence, option D is correct.

Suggest Corrections

Similar questions

2 N a_{2} S_{2} O_{3} + I_{2} \to 2 N a I + N a_{2} S_{4} O_{6}

(N a_{2} S_{2} O_{3} \cdot 5 H_{2} O)

2 N a_{2} S_{2} O_{3} + I_{2} \to 2 N a I + N a_{2} S_{4} O_{6}

I_{2} + 2 N a_{2} S_{2} O_{3} \to 2 N a I + N a_{2} S_{4} O_{6}

I_{2} + 2 N a_{2} S_{2} O_{3} \to 2 N a I + N a_{2} S_{4} O_{6}

2 N a_{2} S_{2} O_{3} + I_{2} = N a_{2} S_{4} O_{6} + 2 N a I, I_{2}

Join BYJU'S Learning Program