Chapter 15, Problem 24A

Interpretation Introduction

Interpretation:

Molarity of iodine in the solution have to be calculated.

Concept Introduction:

Composition of a solution can be defined by expressing its concentration. The concentrations of solutions can be expressed in different ways, which are involved in the quantity of solute and the quantity of solution or solvent. Various methods are used to describe the concentration of the solution quantitatively. Some commonly used quantitative concentration terms are percent by mass, percent by volume, molarity, molality and mole fraction.

Molarity: Molarity is defined as the number of moles of solute present in one litre of the solution In expression, $\mathrm{Molarity}ofthesolution=\frac{Numberofmolesof the solute}{ Volume of the solution(inlitre)}$

Again $\mathrm{Number}ofmolesofsolute=\frac{Massofthesolute}{Molecularweightofthesolute}$

Answer to Problem 24A

Molarity of iodine in the solution is 0.0902 M.

Explanation of Solution

  $\mathrm{Molarity}ofthesolution=\frac{Numberofmolesof the solute}{ Volume of the solution(inlitre)}$

Data given: Mass of I₂= 5.15 gm

Volume of I₂solution = 225 mL = 0.225 L

  $\begin{array}{l}Numberofmolesof{I}_2=\frac{Massof{I}_2}{Molecularweightof{I}_2}\\ =\frac{5.15gm}{253.81gm/mol}=0.0203mol\end{array}$

  $\begin{array}{l}Molarityof{I}_{2}solution=\frac{Numberofmolesof{I}_2}{ Volume of{I}_{2}solution(inlitre)}\\ =\frac{0.0203mole}{ 0.225litre}=0.0902M\end{array}$