Chapter 14.2, Problem 21PP

Interpretation Introduction

Interpretation:

Amount of CaCl₂ is to be calculated to prepare 500 mL of 0.20 M CaCl₂ solution.

Concept introduction:

The concentrations of solutions are expressed in different ways, all of them involved in the quantity of solute and the quantity of solution or solvent. Among them, molarity is widely used to express the concentration of solution. Molarity: Molarity is defined as the number of moles of solute present in one litre of the solution In expression, $\mathrm{Molarity}ofsolution=\frac{Numberofmolesofsolute}{Volumeofsolution(inlitre)}$

Again $\mathrm{Number}ofmolesofsolute=\frac{Massofsolute}{molecularweightofsolute}$

Answer to Problem 21PP

11.1 gm of CaCl₂ is needed to make 500 mL of 0.20 M CaCl₂ solution.

Given information:

Molarity of $CaC{l}_2$ solution = 0.20 M

Volume of $CaC{l}_2$ solution = 500 mL = 0.5 L

Explanation:

Write the expression for the molarity of solution:

$\mathrm{Molarity}ofsolution=\frac{Numberofmolesofsolute}{Volumeofsolution(inlitre)}$

$\mathrm{Molarity}ofCaC{l}_{2}solution=\frac{NumberofmolesofCaC{l}_2}{VolumeofCaC{l}_2}$

Substituting the values:

Number of moles of $CaC{l}_2$ = $0.20M\times 0.5L$

$\begin{array}{l}=0.20M\times 0.5L \\ =0.20\left(mol/litre\right)\times 0.5litre\\ =0.10mol\end{array}$

Again $\mathrm{Number}ofmolesofsolute=\frac{Massofsolute}{\mathrm{Mo}\mathrm{lecular}weightofsolute}$

$\mathrm{Number}ofmolesofCaC{l}_{2}solution=\frac{MassofCaC{l}_2}{MolecularweightofCaC{l}_2}$ = 0.10

$\begin{array}{c} Masso{f}_{ }CaC{l}_2 =110.98\left(gm/mol\right)\times 0.10mol\left(MolecularweightofCaC{l}_2=110.98gm/mol\right)\\ =11.1gm\end{array}$

Thus, the mass of $CaC{l}_2$ is $11.1gm$

Explanation of Solution

Given information:

Molarity of $CaC{l}_2$ solution = 0.20 M

Volume of $CaC{l}_2$ solution = 500 mL = 0.5 L

Write the expression for the molarity of solution:

$\mathrm{Molarity}ofsolution=\frac{Numberofmolesofsolute}{Volumeofsolution(inlitre)}$

$\mathrm{Molarity}ofCaC{l}_{2}solution=\frac{NumberofmolesofCaC{l}_2}{VolumeofCaC{l}_2}$

Substituting the values:

Number of moles of $CaC{l}_2$ = $0.20M\times 0.5L$

$\begin{array}{l}=0.20M\times 0.5L \\ =0.20\left(mol/litre\right)\times 0.5litre\\ =0.10mol\end{array}$

Again $\mathrm{Number}ofmolesofsolute=\frac{Massofsolute}{\mathrm{Mo}\mathrm{lecular}weightofsolute}$

$\mathrm{Number}ofmolesofCaC{l}_{2}solution=\frac{MassofCaC{l}_2}{MolecularweightofCaC{l}_2}$ = 0.10

$\begin{array}{c} Masso{f}_{ }CaC{l}_2 =110.98\left(gm/mol\right)\times 0.10mol\left(MolecularweightofCaC{l}_2=110.98gm/mol\right)\\ =11.1gm\end{array}$

Thus, the mass of $CaC{l}_2$ is $11.1gm$