Chapter 14.2, Problem 17PP

Interpretation Introduction

Interpretation:

Molarity of an aqueous solution of $1.55gmKBr$ in $1.60L$ solution is to be calculated.

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 17PP

Molarity of an aqueous solution of $1.55gmKBr$ in $1.60L$ solution is $8.14\times {10}^{-3}M.$

Explanation:

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

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

Again Number of moles of $KBr$

$\text{=}\frac{\text{Mass}\text{of}\text{KBr}}{\text{Molecular}\text{weight}\text{of}\text{KBr}}$

Data given: Mass of $KBr$ = $1.55gm$

Volume of solution = $1.60L$

Number of moles of $KBr$ = $=\frac{1.15}{1.99}$ (Molecular weight of $KBr$ = $119.0gm/mol$ )

= $0.013$

So, molarity of 0.128 mole $KBr$ solution $\text{=}\frac{\text{0}\text{.013}}{\text{1}\text{.6}}$

= $8.14\times {10}^{-3}M.$

Molarity of an aqueous solution of $1.55gmKBr$ in $1.60L$ solution is $8.14\times {10}^{-3}M.$

Explanation of Solution

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

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

Again Number of moles of $KBr$

$\text{=}\frac{\text{Mass}\text{of}\text{KBr}}{\text{Molecular}\text{weight}\text{of}\text{KBr}}$

Data given: Mass of $KBr$ = $1.55gm$

Volume of solution = $1.60L$

Number of moles of $KBr$ = $=\frac{1.15}{1.99}$ (Molecular weight of $KBr$ = $119.0gm/mol$ )

= $0.013$

So, molarity of 0.128 mole $KBr$ solution $\text{=}\frac{\text{0}\text{.013}}{\text{1}\text{.6}}$

= $8.14\times {10}^{-3}M.$

Conclusion

Molarity of an aqueous solution of $1.55gmKBr$ in $1.60L$ solution is $8.14\times {10}^{-3}M.$