Sodium metal is produced commercially by the electrolysis of molten sodium chloride and chlorine is produced as a by product. How many liters of chlorine at 1.8 atm and ${27}^{o}C$ will be produced if a current of $1.0\times {10}^3$A is passed through $NaCl$ for 9.65 h?

The correct option is C 2463
Given: $I=1.0\times {10}^{3}A$

$t=9.65h=9.65\times 60\times 60s$

$E_{Cl}=35.5$

$W=\frac{ItE}{96500}$

$=\frac{1.0\times {10}^3\times 9.65\times 60\times 60\times 35.5}{96500}$

$=12.78\times {10}^{3}g$

For $C{l}_2\to 71g$ occupies $22.4L$

$\therefore 12.78\times {10}^{3}g$ occupies $\frac{12.78\times {10}^3\times 22.4}{71}=4032L$

$\Rightarrow \frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

$\Rightarrow \frac{1\times 4032}{273}=\frac{1.8\times V_2}{300}$

$\Rightarrow V_2=2461L$

Cell reaction is $2NaCl\to 2N{a}_{\left(s\right)}+{C{l}_2}_{\left(g\right)}$

$\therefore$ Volume of chlorine produced= $2461L$