A saturated solution of potassium chlorate (\text{KClO}_3) at 20^\circ\text{C} contains 7.… — JAMB UTME

A saturated solution of potassium chlorate ( $\text{KClO}_3$ ) at $20^\circ\text{C}$ contains $7.35\text{ g}$ of solute dissolved in $100\text{ g}$ of distilled water. What is the solubility of $\text{KClO}_3$ at $20^\circ\text{C}$ in $\text{mol/dm}^3$ ? [Molar masses: $\text{K} = 39, \text{Cl} = 35.5, \text{O} = 16$ ; density of water $= 1.00\text{ g/cm}^3$ ]

Answer: 0.6 mol/dm³

Answer

The solubility of potassium chlorate at

20^\circ\text{C}

0.6\text{ mol/dm}^3

To convert mass of salt in a given solvent volume into solubility in

\text{mol/dm}^3

, first calculate the molar mass of

\text{KClO}_3

(

122.5\text{ g/mol}

). The amount of salt in moles is

7.35 / 122.5 = 0.06\text{ mol}

. Since

100\text{ g}

of water equals

0.1\text{ dm}^3

, the concentration of the saturated solution is

0.06\text{ mol} / 0.1\text{ dm}^3 = 0.6\text{ mol/dm}^3

Step-by-Step Solution

Calculate the molar mass of

\text{KClO}_3

122.5\text{ g/mol}

Sum the relative atomic masses:

39 (\text{K}) + 35.5 (\text{Cl}) + 3 \times 16 (\text{O}) = 122.5\text{ g/mol}

Calculate the number of moles of solute

0.06\text{ mol}

Divide the given mass by the molar mass:

\frac{7.35\text{ g}}{122.5\text{ g/mol}} = 0.06\text{ mol}

Convert the mass of solvent to volume in

\text{dm}^3

0.1\text{ dm}^3

Water density is

1.00\text{ g/cm}^3

, so

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

Determine the molar solubility

0.6\text{ mol/dm}^3

Divide moles of solute by volume of solvent in

\text{dm}^3

\frac{0.06\text{ mol}}{0.1\text{ dm}^3} = 0.6\text{ mol/dm}^3

Key Concept

Solubility in mol/dm³

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