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(A) The solution X is unsaturated
(B) The solution Y should contain 2.9 g solute in 1000 mL solution.
(C) The percentage by mass of the solute is 40%
(A) The type of solution X is determined as follows;
The maximum amount of solute that can dissolve in solution X = 29 g/Liter
The amount of solute in solution X = 2.9 g/Liter
This solution is unsaturated because the amount of solute present is less than the maximum the solution can take.
(B) The amount solute and volume in the solution Y is calculated as follows;
The solubility = 2.9 g/Liter
[tex]solubility = \frac{2.9 \ g}{L} \times \frac{L}{ mL \times 1000} = \frac{2.9 \ g}{1000mL}[/tex]
Thus, the solution should contain 2.9 g solute in 1000 mL solution.
(C) The percentage by mass of the solute is calculated as;
[tex]mass \ \% = \frac{20 \ g}{50 \ g} \times 100 \% = 40 \%[/tex]
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