Question

If $1.202g{mL}^{-1}$ is the density of $20\mathrm{\%}$ aqueous KI, determine the molality of KI.

• (A) 1.51 m
• (B) 2.53 m
• (C) 3.54 m
• (D) 2.55 m

Answer 1

The Correct Option is A

Explanation:
Molar mass of $KI=39+127=166g{mol}^{-1}$$20\mathrm{\%}$ aqueous solution of KI means $20g$ of $KI$ is present in $100g$ of solution.Therefore, Mass of $KI=20g$That is, $20g$ of $KI$ is present in $(100-20)g$ of water $=80g$ of waterWe know that:Molality of the solution $=\frac{\text{ Moles of }KI}{\text{ Mass of water in }kg}$$=\frac{\frac{\text{Mass of KI}}{\text{Molar Mass of KI}}}{\text{ Mass of water in }kg}$$=\frac{\frac{20}{166}}{0.08}m$$=1.506m$$=1.51m$Hence, the correct option is (A).