Solve the following solubility and crystallization problem. What is the mass of the salt t… — JAMB UTME

Solve the following solubility and crystallization problem. What is the mass of the salt that crystallizes out of the solution upon cooling?

Cevap:A saturated solution of a divalent metal salt

\text{MX}_2

(molar mass =

120\text{ g/mol}

) has a total mass of

120.0\text{ g}

60^\circ\text{C}

. The solubility of

\text{MX}_2

5.0\text{ mol/dm}^3

60^\circ\text{C}

and

2.0\text{ mol/dm}^3

25^\circ\text{C}

. Assuming the density of water is

1.00\text{ g/cm}^3

, the mass of

\text{MX}_2

deposited when the solution is cooled from

60^\circ\text{C}

25^\circ\text{C}

is 【27】 g.

Cevap

The mass of

\text{MX}_2

deposited on cooling is

27.0\text{ g}

60^\circ\text{C}

, a solubility of

5.0\text{ mol/dm}^3

corresponds to

5.0 \times 120 = 600\text{ g}

\text{MX}_2

per

1000\text{ g}

of water. Thus,

1600\text{ g}

of saturated solution contains

600\text{ g}

solute and

1000\text{ g}

water. Proportionally,

120.0\text{ g}

of saturated solution contains

45.0\text{ g}

\text{MX}_2

dissolved in

75.0\text{ g}

of water. At

25^\circ\text{C}

, the solubility decreases to

2.0\text{ mol/dm}^3

, which equals

2.0 \times 120 = 240\text{ g}

\text{MX}_2

per

1000\text{ g}

of water. In

75.0\text{ g}

of water, the maximum mass of solute that remains dissolved is

75.0 \times (240 / 1000) = 18.0\text{ g}

. The mass of solid

\text{MX}_2

that crystallizes out is

45.0\text{ g} - 18.0\text{ g} = 27.0\text{ g}

Adım Adım Çözüm

Calculate the mass of solute per

1000\text{ g}

of water at

60^\circ\text{C}

Mass of

\text{MX}_2

per

1000\text{ g}

water =

5.0\text{ mol/dm}^3 \times 120\text{ g/mol} = 600.0\text{ g}

Converting solubility from

\text{mol/dm}^3

\text{g/dm}^3

(or grams per

1000\text{ g}

of water, given water density is

1.00\text{ g/cm}^3

Determine the composition of the

120.0\text{ g}

saturated solution at

60^\circ\text{C}

Total mass of saturated solution containing

1000\text{ g}

water =

1000\text{ g} + 600\text{ g} = 1600.0\text{ g}

. Mass of water in

120.0\text{ g}

solution =

120.0\text{ g} \times \frac{1000}{1600} = 75.0\text{ g}

. Mass of

\text{MX}_2

dissolved =

120.0\text{ g} - 75.0\text{ g} = 45.0\text{ g}

To find how much solute and solvent are actually present in the given portion of solution.

Calculate the mass of solute that remains dissolved in

75.0\text{ g}

of water at

25^\circ\text{C}

25^\circ\text{C}

, mass of solute per

1000\text{ g}

water =

2.0\text{ mol/dm}^3 \times 120\text{ g/mol} = 240.0\text{ g}

. Mass dissolved in

75.0\text{ g}

water =

75.0\text{ g} \times \frac{240.0}{1000} = 18.0\text{ g}

Determining the maximum amount of solute that

75.0\text{ g}

of water can hold at the lower temperature.

Calculate the mass of

\text{MX}_2

deposited upon cooling.

Mass deposited =

45.0\text{ g} - 18.0\text{ g} = 27.0\text{ g}

The difference between the initial mass dissolved at

60^\circ\text{C}

and the remaining mass dissolved at

25^\circ\text{C}

represents the crystallized solid.

Anahtar Kavram

Quantitative solubility calculations involving crystallization from saturated solutions

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