The equivalent weight of chlorine molecule in the equation 3Cl₂ + 6NaOH → 5NaCl+NaClO₃ + 3H₂O

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

59.1

35.5

42.6

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\large 3\mathop {C{l_2}}\limits^0 + {\text{ }}6NaOH{\text{ }} \to {\text{ }}5Na\mathop {Cl}\limits^{ - 1} + Na\mathop {Cl}\limits^{ + 5} {O_3} + {\text{ }}3{H_2}O$
It is a diproprotionation reaction
Cl₂ in both Oxidant and reductant
$\large \left. {\begin{array}{*{20}{c}} {\mathop {C{l_2}}\limits_0 \; \to \;\mathop {ClO_3^ - }\limits_{ + 5} } \\ {Increase\;in\;Ox.state\; = \;5} \\ {Total\;increase\;inOx.state\; = \;5 \times 2 = 10} \\ {EW\;of\;C{l_2}(reduc\tan t)\; = \;\frac{M}{{10}}} \end{array}} \right|\begin{array}{*{20}{c}} {\mathop {C{l_2}}\limits_0 \; \to \;\mathop {C{l^ - }}\limits_{ - 1} } \\ {Decrease\;in\;Ox.state\; = \;1} \\ {Total\;decrease\;inOx.state\; = \;2 \times 1\; = 2} \\ {EW\;of\;C{l_2}(oxidation)\; = \;\frac{M}{2}} \end{array}$

$\large Ew\;of\;C{l_2}\;in\;the\;reaction\; = \;\frac{M}{{10}}\; + \;\frac{M}{2}$
                                                                 $\large =\;\frac{6M}{10}$
                                                                 $\large =\;\frac{6(71)}{10}$
                                                                 = 42.6