Calculate the solubility in \text{mol/dm}^3 of sodium nitrate (\text{NaNO}_3) at 25^\circ\… — JAMB UTME

Calculate the solubility in $\text{mol/dm}^3$ of sodium nitrate ( $\text{NaNO}_3$ ) at $25^\circ\text{C}$ , if $17.0\text{ g}$ of the salt dissolves in $100.0\text{ g}$ of water to form a saturated solution. [Relative atomic masses: $\text{Na} = 23, \text{N} = 14, \text{O} = 16$ ; density of water $= 1.0\text{ g/cm}^3$ ]

Answer: 2 mol/dm^3

Answer

The solubility of sodium nitrate at

25^\circ\text{C}

2.0\text{ mol/dm}^3

To find solubility in

\text{mol/dm}^3

, first convert

17.0\text{ g}

\text{NaNO}_3

into moles by dividing by its molar mass (

85.0\text{ g/mol}

), obtaining

0.20\text{ mol}

. Next, convert

100.0\text{ g}

of water into volume, which equals

0.100\text{ dm}^3

. Dividing

0.20\text{ mol}

0.100\text{ dm}^3

yields

2.0\text{ mol/dm}^3

Step-by-Step Solution

Determine the molar mass of the solute (

\text{NaNO}_3

)

Molar mass of

\text{NaNO}_3 = 23 + 14 + (3 \times 16) = 85.0\text{ g/mol}

Molar mass is required to convert mass of solute to amount in moles.

Calculate the amount of

\text{NaNO}_3

in moles

\text{Moles} = \frac{17.0\text{ g}}{85.0\text{ g/mol}} = 0.20\text{ mol}

Solubility in

\text{mol/dm}^3

measures the amount of solute in moles per cubic decimeter of solvent.

Convert the mass of solvent (water) to volume in

\text{dm}^3

100.0\text{ g of water} = 100.0\text{ cm}^3 = 0.100\text{ dm}^3

Density of water is

1.0\text{ g/cm}^3

, and

1000\text{ cm}^3 = 1\text{ dm}^3

Calculate the solubility in

\text{mol/dm}^3

\text{Solubility} = \frac{0.20\text{ mol}}{0.100\text{ dm}^3} = 2.0\text{ mol/dm}^3

Dividing the moles of solute by the volume of solvent in

\text{dm}^3

gives the concentration in

\text{mol/dm}^3

Key Concept

Solubility expressed in molar concentration (mol/dm³)

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