At 50^\circ\text{C}, 21.2\text{ g} of anhydrous sodium trioxocarbonate(IV), \text{Na}_2\te… — JAMB UTME

At $50^\circ\text{C}$ , $21.2\text{ g}$ of anhydrous sodium trioxocarbonate(IV), $\text{Na}_2\text{CO}_3$ , is dissolved in $200\text{ cm}^3$ of distilled water to form a saturated solution. Calculate the solubility of the salt at this temperature in $\text{mol/dm}^3$ . $[\text{Na} = 23, \text{C} = 12, \text{O} = 16]$

Answer:The solubility of

\text{Na}_2\text{CO}_3

50^\circ\text{C}

is 【1.00】

\text{mol/dm}^3

Answer

The solubility of

\text{Na}_2\text{CO}_3

50^\circ\text{C}

1.00\text{ mol/dm}^3

To find the solubility in

\text{mol/dm}^3

, first calculate the molar mass of

\text{Na}_2\text{CO}_3

2(23) + 12 + 3(16) = 106\text{ g/mol}

. Next, convert

21.2\text{ g}

of salt into moles:

21.2 / 106 = 0.20\text{ mol}

. Then, convert

200\text{ cm}^3

of solvent into

\text{dm}^3

200 / 1000 = 0.20\text{ dm}^3

. Dividing moles by volume in

\text{dm}^3

gives

0.20 / 0.20 = 1.00\text{ mol/dm}^3

Step-by-Step Solution

Calculate the molar mass of

\text{Na}_2\text{CO}_3

2(23) + 12 + 3(16) = 106\text{ g/mol}

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

Calculate the number of moles of

\text{Na}_2\text{CO}_3

dissolved

\frac{21.2\text{ g}}{106\text{ g/mol}} = 0.20\text{ mol}

Determines the mole quantity of solute present in solution.

Convert the volume of water from

\text{cm}^3

\text{dm}^3

\frac{200\text{ cm}^3}{1000} = 0.20\text{ dm}^3

Solubility in molarity requires volume in cubic decimeters.

Calculate solubility in

\text{mol/dm}^3

\frac{0.20\text{ mol}}{0.20\text{ dm}^3} = 1.00\text{ mol/dm}^3

Solubility is the number of moles of solute per

\text{dm}^3

of solvent.

Key Concept

Solubility Calculation in Moles per Cubic Decimeter

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