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

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59.1

35.5

42.6

answer is A.

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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