Chapter 14, Problem 71A

Interpretation Introduction

Interpretation:

Molarity of CaCO₃ solution have to be calculated.

Concept Introduction:

The 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 liter of the solution.

In expression, $\mathrm{Molarity}\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}solution=\frac{Numberofmolesof the solute}{ Volume of the solution(inlitre)}$

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

Answer to Problem 71A

The molarity of CaCO₃ solution is 0.571.

Explanation of Solution

  $\mathrm{Molarity}\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}solution=\frac{Numberofmolesof the solute}{ Volume of the solution(inlitre)}$

Data given: Mass of CaCO₃ = 15.7 gm

CaCO₃ is dissolved in 275 mL water. So, volume of CaCO₃ solution = 275 mL = 0.275 L

  $\begin{array}{l}Numberofmolesofsolute=\frac{Massofthesolute}{Molecularweightofthesolute}\end{array}$

  $\begin{array}{c}NumberofmolesofCaC{O}_3=\frac{MassofCaC{O}_3}{MolecularweightofCaC{O}_3}\end{array}$

  $\begin{array}{c}NumberofmolesofCaC{O}_3=\frac{15.7gm}{100.01gm/mol}=0.157mole\end{array}$

(Note: Molecular weight of CaCO₃ = 100.01 gm/mol)

  $\begin{array}{l}Molarity\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}solution=\frac{Numberofmolesof the solute}{ Volume of the solution(inlitre)}\end{array}$

  $\begin{array}{c}Molarity\hspace{0.17em}of\hspace{0.17em}CaC{O}_{3}solution=\frac{NumberofmolesofCaC{O}_3}{ Volume of CaC{O}_{3}solution(inlitre)}\end{array}$

  $\begin{array}{c}Molarity\hspace{0.17em}of\hspace{0.17em}CaC{O}_{3}solution=\frac{0.157mole}{ 0.275litre}=0.571M\end{array}$