Suppose that you have made a saturated solution of this solute using 56.0 g of water at 20.0 Â°c. how much more solute can you add if the temperature is increased to 30.0 Â°c?

Mass of water = 56 grams.
So at 20 degrees, Solubility of water = 37.4 g in 100g
Solute = 56 x (37.4 / 100) = 20.944
So at 30 degrees, Solubility of water = 71.5 g in 100g
Solute = 56 x (71.5 / 100) = 40.04
So with the increase of temperature increase in solute = 40.04 - 20.944 = 19
.096 grams

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abhishekprakash2625 abhishekprakash2625 · Answer 2 · 2021-11-22T17:27:30+01:00

The solubility define as the mixing of the solvent into the solution at constant temperature is called solubility. The solubility of the different compounds is different on the basis of their chemical nature.

The solubility depends on these various factors:-

Temperature

Pressure

19.096 grams more solute is required.

The mass of water = 56 grams, So at 20 degrees, The solubility of water will be 37.4 g in 100g

[tex]Solute = 56 \ * \ (\frac{37.4}{100}) = 20.944[/tex]

So at 30 degrees, Solubility of water = 71.5 g in 100g

[tex]Solute = 56\ x \ (\frac{71.5}{ 100}) = 40.04[/tex]

So with the increase of temperature increase in solute = [tex]40.04 - 20.944 = 19.096 grams[/tex]

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