Chapter 14.2, Problem 34SSC

Interpretation Introduction

Interpretation:

Amount of ${\mathrm{NH}}_{4}Cl$ solution is to be calculated to prepare 2.5 L of 0.5 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 34SSC

Amount of ${\mathrm{NH}}_{4}Cl$ solution to prepare 2.5 L of 0.5 M solution is = 66.86 g

Given information:

Molarity of ${\mathrm{NH}}_{4}Cl$ solution = 0.5 M

Volume of ${\mathrm{NH}}_{4}Cl$ solution = 2.5 L

Explanation:

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

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

Number of moles of ${\mathrm{NH}}_{4}Cl$ = 0. 5 M × 2.5 L

= 1.25

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

$=\frac{MassofN{H}_{4}Cl}{MolecularweightofN{H}_{4}Cl}$

= 1.25

That is, Mass of ${\mathrm{NH}}_{4}Cl$ = 53.491  (gm/mol) × 1.25 mol (Molecular weight of ${\mathrm{NH}}_{4}Cl$ = 53.491 gm/mol)

= 66.86 g

Thus, Amount of ${\mathrm{NH}}_{4}Cl$ solution to prepare 2.5 L of 0.5 M solution is = 66.86 g

Explanation of Solution

Given information:

Molarity of ${\mathrm{NH}}_{4}Cl$ solution = 0.5 M

Volume of ${\mathrm{NH}}_{4}Cl$ solution = 2.5 L

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

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

Number of moles of ${\mathrm{NH}}_{4}Cl$ = 0. 5 M × 2.5 L

= 1.25

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

$=\frac{MassofN{H}_{4}Cl}{MolecularweightofN{H}_{4}Cl}$

= 1.25

That is, Mass of ${\mathrm{NH}}_{4}Cl$ = 53.491  (gm/mol) × 1.25 mol (Molecular weight of ${\mathrm{NH}}_{4}Cl$ = 53.491 gm/mol)

= 66.86 g

Thus, Amount of ${\mathrm{NH}}_{4}Cl$ solution to prepare 2.5 L of 0.5 M solution is = 66.86 g