Solubility (g/100gH2O)
Substance 20∘C 50∘C
KCl 34.0 42.6
NaNO3 88.0 114.0
C12H22O11 (sugar) 203.9 260.4
Using the above table, determine whether each of the following solutions will be saturated or unsaturated at 50 ∘C.
1.) Adding 19.7 g KCl to 35.0 g H2O
2.)Adding 184 g NaNO3 to 75.0 g H2O
3.)Adding 87.8 g sugar to 25.0 g H2O
For KCl at 50 C.
42.6 g KCl x (35.0g H2O/100 g H2O) = 14.9 which means 14.9 g KCl can be dissolved in 35.0 g H2O. That means it will be saturated since 19.7 g KCl is greater than what will dissolve.
2 and 3 are done the same way.
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To determine whether each solution will be saturated or unsaturated at 50 °C, we need to compare the amount of solute dissolved (in grams) to the solubility of each substance in 50 g of water at that temperature.
1.) Adding 19.7 g KCl to 35.0 g H2O:
The solubility of KCl in 50 g H2O at 50 °C is 42.6 g/100gH2O.
By using a proportion, we can determine the amount of KCl that can dissolve in 35.0 g H2O:
(42.6 g/100gH2O) = (x/35.0 g)
Cross-multiplying gives us:
x = (42.6 g/100gH2O) * 35.0 g
x ≈ 14.91 g
Since we are adding only 19.7 g of KCl to the solution, which is less than 14.91 g (the maximum amount that can be dissolved in 35.0 g H2O at 50 °C), the solution will be unsaturated.
2.) Adding 184 g NaNO3 to 75.0 g H2O:
The solubility of NaNO3 in 50 g H2O at 50 °C is 114.0 g/100gH2O.
By using a proportion, we can determine the amount of NaNO3 that can dissolve in 75.0 g H2O:
(114.0 g/100gH2O) = (x/75.0 g)
Cross-multiplying gives us:
x = (114.0 g/100gH2O) * 75.0 g
x ≈ 85.5 g
Since we are adding 184 g of NaNO3 to the solution, which is more than 85.5 g (the maximum amount that can be dissolved in 75.0 g H2O at 50 °C), the solution will be saturated.
3.) Adding 87.8 g sugar to 25.0 g H2O:
The solubility of sugar in 50 g H2O at 50 °C is 260.4 g/100gH2O.
By using a proportion, we can determine the amount of sugar that can dissolve in 25.0 g H2O:
(260.4 g/100gH2O) = (x/25.0 g)
Cross-multiplying gives us:
x = (260.4 g/100gH2O) * 25.0 g
x ≈ 65.1 g
Since we are adding 87.8 g of sugar to the solution, which is more than 65.1 g (the maximum amount that can be dissolved in 25.0 g H2O at 50 °C), the solution will be saturated.
To determine whether a solution will be saturated or unsaturated at 50 °C, we need to compare the solubility of the substance at that temperature to the amount of the substance added to the water.
Let's go through each case one by one:
1.) Adding 19.7 g KCl to 35.0 g H2O:
From the table, we see that at 50 °C, the solubility of KCl is 42.6 g/100g H2O. We need to compare this to the amount of KCl added, which is 19.7 g. We can calculate the solubility of the solution by dividing the amount of KCl added (19.7 g) by the amount of water (35.0 g) and multiplying by 100.
Solubility of the solution = (19.7 g / 35.0 g) * 100 = 56.3 g/100g H2O.
Since the solubility of the solution (56.3 g/100g H2O) is greater than the solubility of KCl at 50 °C (42.6 g/100g H2O), the solution is saturated.
2.) Adding 184 g NaNO3 to 75.0 g H2O:
From the table, we see that at 50 °C, the solubility of NaNO3 is 114.0 g/100g H2O. We can again calculate the solubility of the solution by dividing the amount of NaNO3 added (184 g) by the amount of water (75.0 g) and multiplying by 100.
Solubility of the solution = (184 g / 75.0 g) * 100 = 245.3 g/100g H2O.
Since the solubility of the solution (245.3 g/100g H2O) is greater than the solubility of NaNO3 at 50 °C (114.0 g/100g H2O), the solution is saturated.
3.) Adding 87.8 g sugar to 25.0 g H2O:
From the table, we see that at 50 °C, the solubility of sugar (C12H22O11) is 260.4 g/100g H2O. We can calculate the solubility of the solution by dividing the amount of sugar added (87.8 g) by the amount of water (25.0 g) and multiplying by 100.
Solubility of the solution = (87.8 g / 25.0 g) * 100 = 351.2 g/100g H2O.
Since the solubility of the solution (351.2 g/100g H2O) is greater than the solubility of sugar at 50 °C (260.4 g/100g H2O), the solution is saturated.
Therefore, all three solutions will be saturated at 50 °C.