The solubility of sodium selenate is 84 g/100 g of water at 35 oC. If a solution is obtained by dissolving 92 g of sodium selenate in 200 g of water at 35 oC, What do you call this solution?
(A) Diluted
(B) Saturated
(C) Unsaturated
(D) supersaturated
92g/200 is the same as 46g/100. Since the solubility is 84/100, it certainly can't be saturated or supersaturated. So it must be .......
Answer
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Give the answer of the given example
To determine the type of solution formed when dissolving 92 g of sodium selenate in 200 g of water at 35 °C, we need to compare the amount of solute (sodium selenate) dissolved to the solubility of sodium selenate at that temperature.
The solubility of sodium selenate is given as 84 g/100 g of water at 35 °C. This means that at 35 °C, 100 g of water can dissolve 84 g of sodium selenate.
In the given situation, we have 200 g of water and 92 g of sodium selenate. To find out whether this solution is diluted, saturated, unsaturated, or supersaturated, we need to determine how much sodium selenate can dissolve in 200 g of water at 35 °C.
Using a proportion, we can calculate the expected solubility of sodium selenate in 200 g of water:
(84 g/100 g) = (x g/200 g)
Solving for x:
x = (84 g/100 g) * 200 g
x = 168 g
The expected solubility of sodium selenate in 200 g of water at 35 °C is 168 g. However, we have dissolved 92 g of sodium selenate in 200 g of water, which is less than the expected solubility.
Therefore, the solution is unsaturated because the amount of solute dissolved (92 g) is less than the solubility of sodium selenate in 200 g of water (168 g) at 35 °C.
So the correct answer is (C) Unsaturated.