Question

How much $PC{l}_5$ must be added to a one litre vessel kept at ${250}^o$C in order to obtain $0.1$ mole of $C{l}_2$ gas?

[$K_C$ for $PC{l}_{5\left(g\right)}\rightleftharpoons PC{l}_{3\left(g\right)}+C{l}_{2\left(g\right)}$ is $0.0414$ mol/L]

• A. $0.0341$ mol
• B. $0.341$ mol
• C. $0.241$ mol
• D. $0.024$ mol

Answer 1

The correct option is B $0.341$ mol

Since the total volume is 1 L, the number of moles is equal to molar concentration.

Let x mole of $PC{l}_5$ is added. The equilibrium concentration of $PC{l}_5$ is $x-0.1$ M as 0.1 mole of $PC{l}_5$ will react to give 0.1 mole of $PC{l}_3$ and 0.1 mole of $C{l}_2$ at equilibrium.

$K_c=\frac{\left[PC{l}_3\right]\left[C{l}_2\right]}{\left[PC{l}_5\right]}$

$0.0414mol/L=\frac{0.1mol/L\times 0.1mol/L}{(x-0.1)mol/L}$

$0.0414(x-0.1)=0.01$

$0.0414x-0.00414=0.01$

$0.0414x=0.01414$

$x=0.341$

Hence, 0.341 mole of $PC{l}_5$ is added.