The following table gives the solubility of different salts at different temperatures.

Temperature (in K)	Substance dissolved (in g)
	KNO3	NaCl	KCl	NH4Cl
283	21	36	35	24
293	32	36	35	37
313	62	36	40	41
333	106	37	46	55
353	167	37	54	66

Answer the following questions based on the table given above.

(a) What mass of KNO₃ would be needed to produce a saturated solution of KNO₃ in 50 grams of water at 313 K.

(b) If a saturated solution of KCl is made at 353 K and then cooled at room temperature, what would you observe? Explain.

(c) Find the solubility of each salt at 293 K.

(d) Which salt has the lowest solubility at 283 K?

(e) What is the effect of change of temperature on the solubility of a salt.

(a) From the table, the solubility of KNO₃ at 313 K is 62 g per 100 g of water.

62 g of KNO₃ per 100 g of water

Mass of solute per 50 g of water is

= $\dfrac{62}{100}$ x 50

= 31 g

∴ Mass of solute per 50 g of water is 31 g.

(b) The solubility of KCl at 353 K is 54 g per 100 g of water.

The solubility of KCl at 293 K (room temperature) is 35 G per 100 g of water.

When the solution cools, the solubility decreases.

The excess KCl will precipitate out of the solution as crystals.

(c) Considering given salts are dissolved in 100g of water

Solubility of KNO₃ at 293 K

Mass of solute = 32 g
Mass of solvent = 100 g

Solubility = $\dfrac{\text{Mass of solute}}{\text{Mass of solvent}}$ x 100

= $\dfrac{32}{100}$ x 100

= 32 g

∴ Solubility of KNO₃ at 293 K is 32 g.

Solubility of NaCl at 293 K

Mass of solute = 36 g
Mass of solvent = 100 g

Solubility = $\dfrac{\text{Mass of solute}}{\text{Mass of solvent}}$ x 100

= $\dfrac{36}{100}$ x 100

= 36 g

∴ Solubility of NaCl at 293 K is 36 g.

Solubility of KCl at 293 K

Mass of solute = 35 g
Mass of solvent = 100 g

Solubility = $\dfrac{\text{Mass of solute}}{\text{Mass of solvent}}$ x 100

= $\dfrac{35}{100}$ x 100

= 35 g

∴ Solubility of KCl at 293 K is 35 g.

Solubility of NH₄Cl at 293 K

Mass of solute = 37 g
Mass of solvent = 100 g

Solubility = $\dfrac{\text{Mass of solute}}{\text{Mass of solvent}}$ x 100

= $\dfrac{37}{100}$ x 100

= 37 g

∴ Solubility of NH₄Cl at 293 K is 37 g.

(d) From the table, when we calculate the solubility of KNO₃ at 283 K,

Mass of solute = 21 g
Mass of solvent = 100 g

Solubility = $\dfrac{\text{Mass of solute}}{\text{Mass of solvent}}$ x 100

= $\dfrac{21}{100}$ x 100

= 21 g

Whereas, the solubility of NaCl, KCl and NH₄Cl will be 36 g, 35 g and 24 g respectively.

Hence, the least solubility will be of KNO₃ at 283 K

(e) The solubility of most salts in water usually increases with rise in temperature. Example, Potassium nitrate.
There are some salts which show anomalous solubility. Their solubility first increases, and then decreases, with rise in temperature.
Example: Na₂SO₄.10H₂O (Glauber's salt).
Solubility curve of Na₂SO₄.10H₂O rises till it reaches 32.8°C, and then it falls slightly. This is because Na₂SO₄.10H₂O is hydrous below 32.8°C and looses it water and become anhydrous above 32.8°C.