At 25^\circ\text{C}, 16.4\text{ g} of anhydrous calcium trioxonitrate(V), \text{Ca(NO}_3\t… — JAMB UTME

At $25^\circ\text{C}$ , $16.4\text{ g}$ of anhydrous calcium trioxonitrate(V), $\text{Ca(NO}_3\text{)}_2$ , is dissolved in $250\text{ cm}^3$ of distilled water to form a saturated solution. What is the solubility of the salt in $\text{mol/dm}^3$ at $25^\circ\text{C}$ ? $[\text{Ca} = 40, \text{N} = 14, \text{O} = 16]$

$0.40\text{ mol/dm}^3$ Answer
B
$65.60\text{ mol/dm}^3$
C
$0.10\text{ mol/dm}^3$
D
$0.64\text{ mol/dm}^3$

Answer

The solubility of calcium trioxonitrate(V) at

25^\circ\text{C}

0.40\text{ mol/dm}^3

The correct answer is obtained by calculating the molar mass of calcium trioxonitrate(V) (

164\text{ g/mol}

), finding the moles dissolved (

0.10\text{ mol}

), and dividing by the solvent volume in cubic decimetres (

0.25\text{ dm}^3

), giving

0.40\text{ mol/dm}^3

Step-by-Step Solution

Calculate the molar mass of

\text{Ca(NO}_3\text{)}_2

\text{Molar Mass} = 40 + 2(14 + 3 \times 16) = 40 + 2(62) = 164\text{ g/mol}

Molar mass is needed to convert the mass of the salt into moles.

Calculate the number of moles of

\text{Ca(NO}_3\text{)}_2

\text{Moles} = \frac{16.4\text{ g}}{164\text{ g/mol}} = 0.10\text{ mol}

Solubility in

\text{mol/dm}^3

requires the amount of solute in moles.

Convert the volume of distilled water from

\text{cm}^3

\text{dm}^3

\text{Volume} = \frac{250\text{ cm}^3}{1000} = 0.25\text{ dm}^3

Concentration units are expressed per cubic decimetre (

\text{dm}^3

Determine the molar solubility

\text{Solubility} = \frac{0.10\text{ mol}}{0.25\text{ dm}^3} = 0.40\text{ mol/dm}^3

Solubility in molar concentration is moles of solute divided by volume of solvent in

\text{dm}^3

Key Concept

Solubility Calculations in Molar Concentration

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