Chapter 14.2, Problem 22PP

Interpretation Introduction

Interpretation:

Amount of NaOHis to be calculated to prepare 250 mL of 3.0 M NaOHsolution.

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.

Answer to Problem 22PP

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

30 gm of NaOHis needed to make 250 mL of 3.0 M NaOHsolution.

Explanation:

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

$\mathrm{Molarity}ofNaOHsolution=\frac{NumberofmolesofNaOH}{VolumeofNaOH}$

Data given: Molarity of NaOHsolution = 3.0 M

Volume of NaOHsolution = 250 mL = 0.25 L

Calculate the number of moles of NaOH

$\begin{array}{l}=3.0M\times 0.25L\hfill \\ \begin{array}{c}=3.0\left(mol/litre\right)\times 0.25litre\\ =0.75mol\end{array}\hfill \end{array}$

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

$\begin{array}{c}\mathrm{Number}ofmolesofNaOHsolution=\frac{MassofNaOH}{MolecularweightofNaOH}\\ =0.75\end{array}$

Calculating the mass of NaOH,

$\begin{array}{l}MassofNaOH=40\left(gm/mol\right)\times 0.75mol(MolecularweightofNaOH=40gm/mol)\hfill \\ =30gm\hfill \end{array}$

Explanation of Solution

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

$\mathrm{Molarity}ofNaOHsolution=\frac{NumberofmolesofNaOH}{VolumeofNaOH}$

Data given: Molarity of NaOHsolution = 3.0 M

Volume of NaOHsolution = 250 mL = 0.25 L

Calculate the number of moles of NaOH

$\begin{array}{l}=3.0M\times 0.25L\hfill \\ \begin{array}{c}=3.0\left(mol/litre\right)\times 0.25litre\\ =0.75mol\end{array}\hfill \end{array}$

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

$\begin{array}{c}\mathrm{Number}ofmolesofNaOHsolution=\frac{MassofNaOH}{MolecularweightofNaOH}\\ =0.75\end{array}$

Calculating the mass of NaOH,

$\begin{array}{l}MassofNaOH=40\left(gm/mol\right)\times 0.75mol(MolecularweightofNaOH=40gm/mol)\hfill \\ =30gm\hfill \end{array}$