Q4 / 155 · WBJEE 2024 · Chemistry

The equivalent weight of $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3($ Gram molecular weight $=M)$ in the given reaction is $\mathrm{I}_2+2 \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3=2 \mathrm{NaI}+\mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6$

  • A
    M/2
  • Correct answer
    B
    M
  • C
    2M
  • D
    M/4

Solution

Hint: $\mathrm{I}_2+2 \mathrm{Na}_2 \stackrel{+2}{\mathrm{~S}}_2 \mathrm{O}_3=2 \mathrm{NaI}+\mathrm{Na}_2 \stackrel{+2.5}{\mathrm{~S}_4} \mathrm{O}_6$ $n \cdot f=2 \times 0.5=1$ $\therefore$ Number of moles of electron lost per mole of $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3=1$ Hence equivalent weight of $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3=\frac{\mathrm{M}}{1}=\mathrm{M}$

Hence, the correct answer is option: 'B'.

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