Chapter 14.2, Problem 19PP

Interpretation Introduction

Interpretation:

Amount of $Ca{\left(OH\right)}_2$ is to be calculated to prepare $1.5L$ of $0.25M$ 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 19PP

$27.78gm$ of $Ca{\left(OH\right)}_2$ is needed to make $1.5L$ of $0.25M$ solution.

Given information:

The molarity of $Ca{\left(OH\right)}_2$ solution = $0.25M$

The Volume of $Ca{\left(OH\right)}_2$ solution = $1.5L$

Explanation:

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

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

Number of moles of $Ca{\left(OH\right)}_2$

$\begin{array}{l}\text{=}\text{0}\text{.25 }\left(\text{mol/litre}\right)\text{ × 1}\text{.5 litre }\\ \text{= 0}\text{.375}\end{array}$

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

$\text{=}\frac{\text{Mass}\text{of}{\text{Ca(OH)}}_2}{\text{Molecular}\text{weight}\text{of}{\text{Ca(OH)}}_2}$

= $0.375$

Molecular weight of $Ca{\left(OH\right)}_2$ = $74.08gm/mol$

Mass of $Ca{\left(OH\right)}_2$ = $74.08\left(gm/mol\right)\times 0.375mol$

= $27.78gm$

$27.78gm$ of $Ca{\left(OH\right)}_2$ is needed to make $1.5L$ of $0.25M$ solution.

Explanation of Solution

Given information:

The molarity of $Ca{\left(OH\right)}_2$ solution = $0.25M$

The Volume of $Ca{\left(OH\right)}_2$ solution = $1.5L$

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

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

Number of moles of $Ca{\left(OH\right)}_2$

$\begin{array}{l}\text{=}\text{0}\text{.25 }\left(\text{mol/litre}\right)\text{ × 1}\text{.5 litre }\\ \text{= 0}\text{.375}\end{array}$

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

$\text{=}\frac{\text{Mass}\text{of}{\text{Ca(OH)}}_2}{\text{Molecular}\text{weight}\text{of}{\text{Ca(OH)}}_2}$

= $0.375$

Molecular weight of $Ca{\left(OH\right)}_2$ = $74.08gm/mol$

Mass of $Ca{\left(OH\right)}_2$ = $74.08\left(gm/mol\right)\times 0.375mol$

= $27.78gm$

Conclusion

$27.78gm$ of $Ca{\left(OH\right)}_2$ is needed to make $1.5L$ of $0.25M$ solution.