A saturated solution of potassium chloride (\text{KCl}) contains 14.9\text{ g} of the salt… — JAMB UTME

A saturated solution of potassium chloride ( $\text{KCl}$ ) contains $14.9\text{ g}$ of the salt dissolved in $100\text{ g}$ of water at $25^\circ\text{C}$ . What is the solubility of potassium chloride in $\text{mol/dm}^3$ at this temperature?

(Take density of water = $1.0\text{ g/cm}^3$ , relative atomic masses: $\text{K} = 39$ , $\text{Cl} = 35.5$ )

Answer: 2 mol/dm³

Answer

The solubility of potassium chloride at

25^\circ\text{C}

2.0\text{ mol/dm}^3

The correct calculated value is

2.0\text{ mol/dm}^3

. Molar mass of

\text{KCl} = 39 + 35.5 = 74.5\text{ g/mol}

. Number of moles of

\text{KCl} = \frac{14.9\text{ g}}{74.5\text{ g/mol}} = 0.2\text{ mol}

. Volume of water solvent

= 100\text{ g} = 0.1\text{ dm}^3

. Therefore, solubility

= \frac{0.2\text{ mol}}{0.1\text{ dm}^3} = 2.0\text{ mol/dm}^3

Step-by-Step Solution

Calculate the molar mass of

\text{KCl}

and convert mass of solute to moles.

Molar mass

= 74.5\text{ g/mol}

; Moles

= \frac{14.9\text{ g}}{74.5\text{ g/mol}} = 0.2\text{ mol}

Solubility in

\text{mol/dm}^3

requires the quantity of solute to be expressed in moles rather than grams.

Convert the mass/volume of solvent into decimeters cubed (

\text{dm}^3

Volume

= 100\text{ g} = 100\text{ cm}^3 = 0.1\text{ dm}^3

Molar solubility concentration is defined per

1.0\text{ dm}^3

of solution/solvent.

Compute the concentration in

\text{mol/dm}^3

by dividing moles of solute by volume of solvent in

\text{dm}^3

Solubility

= \frac{0.2\text{ mol}}{0.1\text{ dm}^3} = 2.0\text{ mol/dm}^3

Solubility in molarity equal to total moles divided by total volume in

\text{dm}^3

Key Concept

Calculating molar solubility in

\text{mol/dm}^3

from solute mass and solvent volume.

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