Question

A current strength of $9.65A$ is passed through excess fused $AlC{l}_3$ for $5h$. How many litres of chlorine will be liberated at $STP$?

$[F=96500C]$

• A. $2.016$
• B. $1.008$
• C. $11.2$
• D. $20.16$
• E. $10.08$

Answer 1

The correct option is D $20.16$

From Faraday first law of electrolysis:

$w=zit$

$z=\frac{\text{Atomic mass}}{\text{electrons exhanged}\times 96500}$

$C{l}^{-}\to \frac{1}{2}C{l}_2+1e^{-}$

$w=\frac{35.5\times 9.65\times 5\times 60\times 60}{1\times 96500}$

$=63.9g$

$\because$ At $STP,71gC{l}_2$ has volume $=22.4L$

$\therefore$ At $STP,63.9gC{l}_2$ will have volume $=\frac{22.4\times 63.9}{71}=20.16L$.

Hence, option $D$ is correct.