Thank you for visiting If 14 2 g of Al NO3 3 is dissolved in 655 g of water what is the boiling point of the solution Assume 100. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
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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