Question

The equivalent weight of chlorine molecule in the equation:
$3C{l}_2+6NaOH\to 5NaCl+NaCl{O}_3+3H_{2}O$

• A. $42.6$
• B. $35.5$
• C. $59.1$
• D. $71$

Answer 1

Answer: (A)

$42.6$

For the relation -

$NaoH+3c{l}_2\to 5Nacl+Nacl{o}_3+3H_{2}O$

In 1st case, $c{l}_2^0\to Nac{l}^{-1}$

O.N. decrease by 1

So, n - factor = 1

Equivalent weight $=\frac{m}{1}...\left(1\right)$

In $2^{nd}case,c{l}_2^0\to Nacl{o}_3$

O.N. increase by 5

so n - factor = 5

Equivalent weight $=\frac{M}{5}...\left(2\right)$

Adding (1) & (2)

$\frac{M}{1}+\frac{M}{5}\Rightarrow \frac{35.5}{1}+\frac{35.5}{5}$

$\Rightarrow 3.5+7.1$

$\Rightarrow 42.6$

So, the correct option is $A$