Question

Calculate $K_C$ for the reaction: $KI(aq.)+I_2\left(s\right)\rightleftharpoons K{I}_3(aq.)$. Given that initial weight of $KI$ is $1.326gm$ and weight of $K{I}_3$ is $0.105gm$ and number of moles of free $I_2$ is $0.0025$ at equilibrium. The volume of solution is $1$ litres. $(M_w$ of $KI=166,M_w$ of $K{I}_3=420)$.

• A. $0.032$
• B. $0.024$
• C. $0.064$
• D. $0.012$

Answer 1

The correct option is A $0.032$

Given

Initial weght of KI is 1.328g

Initial weight of is 0.105g

No, of moles of is o.oo25g

Volume of solution is 1L

Solution

Reaction is

KI(aq)+I_{2}(s)\rightarrow KI_{3}(aq)

As I2 is in solid states moles of KI get transformed into moles of KI3

After some time

Moles of KI left=$\frac{1.326}{166}-\frac{0.105}{420}$=$2.5\ast {10}^{-7}$

$K^c=\frac{\left[K{I}_3\right]}{\left[KI\right]}=2.5\ast {10}^{-7}/7.738\ast {10}^{-3}$=0.032

The correct option is A