At 60^\circ\text{C}, the solubility of a trioxonitrate(V) salt Y (molar mass = 160\text{ g… — JAMB UTME

At $60^\circ\text{C}$ , the solubility of a trioxonitrate(V) salt $Y$ (molar mass $= 160\text{ g mol}^{-1}$ ) is $1.25\text{ mol dm}^{-3}$ . When the solution is cooled to $20^\circ\text{C}$ , its solubility decreases to $0.25\text{ mol dm}^{-3}$ . What mass of salt $Y$ will crystallize out when $300\text{ g}$ of a saturated solution of $Y$ at $60^\circ\text{C}$ is cooled to $20^\circ\text{C}$ ? (Assume the density of water is $1.0\text{ g cm}^{-3}$ ).

$40.0\text{ g}$ Cevap
B
$48.0\text{ g}$
C
$50.0\text{ g}$
D
$160.0\text{ g}$

Cevap

The mass of salt

Y

that crystallizes out upon cooling is

40.0\text{ g}

Converting concentrations to mass per

1000\text{ g}

of water gives

200\text{ g}

60^\circ\text{C}

and

40\text{ g}

20^\circ\text{C}

. A

300\text{ g}

sample of saturated solution at

60^\circ\text{C}

contains

50\text{ g}

of solute dissolved in

250\text{ g}

of solvent (water). Upon cooling to

20^\circ\text{C}

250\text{ g}

of water can only hold

10\text{ g}

of solute. Thus,

50\text{ g} - 10\text{ g} = 40.0\text{ g}

crystallizes out.

Adım Adım Çözüm

Convert solubility at

60^\circ\text{C}

from

\text{mol dm}^{-3}

\text{g dm}^{-3}

\text{Solubility at } 60^\circ\text{C} = 1.25\text{ mol dm}^{-3} \times 160\text{ g mol}^{-1} = 200\text{ g dm}^{-3}

Solubility calculations involving mass of solution require mass concentration (

\text{g dm}^{-3}

Determine mass of water in

300\text{ g}

of saturated solution at

60^\circ\text{C}

\text{Mass of solution per dm}^3 = 1000\text{ g (water)} + 200\text{ g (salt)} = 1200\text{ g}

. Therefore,

\text{mass of water in } 300\text{ g solution} = \frac{1000\text{ g}}{1200\text{ g}} \times 300\text{ g} = 250\text{ g}

Solubility expresses mass of solute per fixed mass of solvent (

1000\text{ g}

water).

Convert solubility at

20^\circ\text{C}

\text{g dm}^{-3}

and find mass of salt remaining dissolved in

250\text{ g}

water

\text{Solubility at } 20^\circ\text{C} = 0.25\text{ mol dm}^{-3} \times 160\text{ g mol}^{-1} = 40\text{ g dm}^{-3}

. In

250\text{ g}

of water, dissolved salt

= 40\text{ g} \times \frac{250\text{ g}}{1000\text{ g}} = 10\text{ g}

Determining how much solute stays dissolved at the lower temperature.

Calculate the mass of salt precipitated

\text{Original salt in solution} = 200\text{ g} \times \frac{250\text{ g}}{1000\text{ g}} = 50\text{ g}

\text{Mass precipitated} = 50\text{ g} - 10\text{ g} = 40.0\text{ g}

Mass precipitated is the difference between initial dissolved mass and remaining dissolved mass.

Anahtar Kavram

Mass of salt precipitated upon cooling saturated solutions

Tahmini Süre:2m 30s

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