Chapter 14.2, Problem 23PP

Interpretation Introduction

Interpretation:

Volume of ethanolis to be calculated to prepare 100 mL of 0.15 M 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 23PP

0.875 mL of ethanolis needed to make 100 mL of 0.15 Msolution.

Given information:

The molarity of ethanolsolution = 0.15 M

The volume of ethanolsolution = 100 mL = 0.1 L

The density of ethanol = 0.7893 gm/mL

Explanation:

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

$\mathrm{Molarity}ofethanolsolution=\frac{Numberofmolesofethanol}{Volumeofethanol}$

Calculating the no. of moles of ethanol:

$\begin{array}{l}Numberofmolesofethanol=Molarityofethanolsolution\times Volumeofethanolsolution\hfill \\ =0.15M\times 0.1L\hfill \\ \begin{array}{c} =0.15\left(mol/litre\right)\times 0.1litre\\ =0.015mol\end{array}\hfill \end{array}$

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

$\mathrm{Number}ofmolesofethanolsolution=\frac{Massofethanol}{Molecularweightofethanol}$

Calculate the mass of ethanol from above equation:

$\begin{array}{c}Massofethanol=46.07\left(gm/mol\right)\times 0.015mol\\ =0.691gm\end{array}$

Calculating the volume of ethanol:

$\begin{array}{c}\mathrm{Volume}ofethanolsolution=\frac{Massofethanol}{Densityofethanol}\\ =\frac{0.691}{0.7893}\\ =0.875mL\end{array}$

Thus, 0.875 mL of ethanolis needed to make 100 mL of 0.15 Msolution.

Explanation of Solution

Given information:

The molarity of ethanolsolution = 0.15 M

The volume of ethanolsolution = 100 mL = 0.1 L

The density of ethanol = 0.7893 gm/mL

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

$\mathrm{Molarity}ofethanolsolution=\frac{Numberofmolesofethanol}{Volumeofethanol}$

Calculating the no. of moles of ethanol:

$\begin{array}{l}Numberofmolesofethanol=Molarityofethanolsolution\times Volumeofethanolsolution\hfill \\ =0.15M\times 0.1L\hfill \\ \begin{array}{c} =0.15\left(mol/litre\right)\times 0.1litre\\ =0.015mol\end{array}\hfill \end{array}$

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

$\mathrm{Number}ofmolesofethanolsolution=\frac{Massofethanol}{Molecularweightofethanol}$

Calculate the mass of ethanol from above equation:

$\begin{array}{c}Massofethanol=46.07\left(gm/mol\right)\times 0.015mol\\ =0.691gm\end{array}$

Calculating the volume of ethanol:

$\begin{array}{c}\mathrm{Volume}ofethanolsolution=\frac{Massofethanol}{Densityofethanol}\\ =\frac{0.691}{0.7893}\\ =0.875mL\end{array}$

Thus, 0.875 mL of ethanolis needed to make 100 mL of 0.15 Msolution.