Answer :
Final answer:
The boiling point of the solution, when 14.2 g of Al(NO3)3 is dissolved in 655g of water, is 102.08 °C.
Explanation:
To find the boiling point of the solution, we need to calculate the boiling point elevation caused by the dissolved Al(NO3)3. The boiling point elevation is given by the equation:
ΔTb = kb * m * i
Where ΔTb is the boiling point elevation, kb is the molal boiling point elevation constant, m is the molality of the solute, and i is the van't Hoff factor.
First, let's calculate the molality of the Al(NO3)3:
Molar mass of Al(NO3)3 = 26.98 g/mol (Al) + 3 * (14.01 g/mol (N) + 16.00 g/mol (O)) = 213.00 g/mol
Moles of Al(NO3)3 = mass / molar mass = 14.2 g / 213.00 g/mol = 0.067 mol
Mass of water = 655 g
Moles of water = mass / molar mass = 655 g / 18.02 g/mol = 36.34 mol
Molality of Al(NO3)3 = moles of solute / mass of solvent (in kg) = 0.067 mol / 0.655 kg = 0.102 mol/kg
Since Al(NO3)3 dissociates into 4 ions in water, the van't Hoff factor (i) is 4.
Now, let's calculate the boiling point elevation:
ΔTb = kb * m * i = 0.52 °C/m * 0.102 mol/kg * 4 = 2.08 °C
The boiling point elevation is 2.08 °C.
Finally, we add the boiling point elevation to the boiling point of pure water (100 °C) to find the boiling point of the solution:
Boiling point of the solution = 100 °C + 2.08 °C = 102.08 °C
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