A solution is a homogeneous mixture of one or more solutes dissolved in a solvent. A specific volume of solvent is only able to
dissolve a limited amount of solute. As long as the solvent is able to dissolve more solute, the solution remains unsaturated. When the solvent can no longer dissolve additional solute, the solution is saturated. At this point, any additional solute will fall to the bottom of the container. The amount of solute that can be dissolved in a given amount of solvent at a specific temperature and pressure is defined as the solubility of the solute.
Consider the solubility curves of several salts in water.
Based on the information here, if 220 grams of the salt KBr are added to 100 mL of water at 100oC, we would label that solution as
A) insoluble
B) saturated
C) supersaturated
D) unsaturated
Consider the solubility curves of several salts in water.
Based on the information here, if 220 grams of the salt KBr are added to 100 mL of water at 100oC, we would label that solution as
A) insoluble
B) saturated
C) supersaturated
D) unsaturated
2 answers:
Answer:
The solution is saturated.
Explanation:
Let´s consider the solubility curve for KBr, which represents the maximum amount of KBr that can be dissolved in 100 g of water at a certain temperature. The solubility of KBr at 100°C is approximately 105 g of KBr every 100 g of water.
The density of water is 1 g/mL. So, 100 g = 100 mL
If we add 220 g of KBr to 100 mL of water, only 105 g of KBr will dissolve and the rest (220 g - 105 g = 115 g) will precipitate. Since the solution has dissolved the maximum amount of solute possible, it is saturated.
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Answer: 4.22 grams of solute is there in 278 ml of 0.038 M
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
where,
n = moles of solute
= volume of solution in L
Now put all the given values in the formula of molality, we get
mass of =
Thus 4.22 grams of solute is there in 278 ml of 0.038 M