Question

How many moles per liter of $PC{l}_5$ has to be taken to obtain $0.1$ mole $C{l}_2$, if the value of equilibrium constant $K_c$ is $0.04$?

• A. $0.15$
• B. $0.25$
• C. $0.35$
• D. $0.05$

Answer 1

Answer: (B)

$0.25$

Consider a as initial concentration of $PC{l}_5$ and x be degree of dissociation.

Dissociation of $PC{l}_5$ takes place according to following reaction

$PC{l}_5\to PC{l}_3+C{l}_2$

Initial: $a$ $0$ $0$

Final: $a(1-x)$ $ax$ $ax$

[$C{l}_2$] $=0.1=ax$ ...(1)

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

$0.04=\frac{(ax.ax)}{a(1-x)}$

Solving this we'll get,

$x=0.4$

From (i)

$a=0.1/0.4=0.25$

Hence, we can say moles of $PC{l}_5$ added is 0.25 moles.