Answer:
a) 0.0265M
b) 8.35×10^-4M
c) 9.833×10^-8M
Explanation:
First we must write down the equation of each reaction. After identifying the insoluble precipitate, we go ahead to write down the expression for its solubility product. If the solubility product of the compound is known, the concentration of the precipitating agent can now be calculated from the available information in the question.
Answer:
34,6g of (NH₄)₂SO₄
Explanation:
The boiling-point elevation describes the phenomenon in which the boiling point of a liquid increases with the addition of a compound. The formula is:
ΔT = kb×m
Where ΔT is Tsolution - T solvent; kb is ebullioscopic constant and m is molality of ions in solution.
For the problem:
ΔT = 109,7°C-108,3°C = 1,4°C
kb = 1.07 °C kg/mol
Solving:
m = 1,31 mol/kg
As mass of X = 600g = 0,600kg:
1,31mol/kg×0,600kg = 0,785 moles of ions. As (NH₄)₂SO₄ has three ions:
0,785 moles of ions× = 0,262 moles of (NH₄)₂SO₄
As molar mass of (NH₄)₂SO₄ is 132,14g/mol:
0,262 moles of (NH₄)₂SO₄× = <em>34,6g of (NH₄)₂SO₄</em>
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I hope it helps!