What is the predicted change in the boiling point of water when 1.50 g of barium chloride (BaCl2) is dissolved in 1.50 kg of water?Kb of water = 0.51°C/molmolar mass BaCl2 = 208.23 g/moli value of BaCl2 = 3A.0.51°CB.-0.00082°CC.0.0073°CD.-0.0024°C
Question
What is the predicted change in the boiling point of water when 1.50 g of barium chloride (BaCl2) is dissolved in 1.50 kg of water?Kb of water = 0.51°C/molmolar mass BaCl2 = 208.23 g/moli value of BaCl2 = 3A.0.51°CB.-0.00082°CC.0.0073°CD.-0.0024°C
Solution
To solve this problem, we need to use the formula for boiling point elevation, which is ΔTb = i * Kb * m, where:
- ΔTb is the change in boiling point
- i is the van 't Hoff factor (number of particles the solute splits into in solution)
- Kb is the boiling point elevation constant
- m is the molality of the solution
Given in the problem, we have:
- i = 3 (for BaCl2, because it dissociates into 3 ions in solution: Ba2+ and 2Cl-)
- Kb = 0.51°C/mol
- molar mass of BaCl2 = 208.23 g/mol
First, we need to calculate the molality (m) of the solution. Molality is defined as the number of moles of solute per kilogram of solvent.
-
Calculate the number of moles of BaCl2:
Number of moles = mass of solute / molar mass Number of moles = 1.50 g / 208.23 g/mol = 0.0072 mol
-
Calculate the molality:
Molality (m) = number of moles / mass of solvent in kg Molality (m) = 0.0072 mol / 1.50 kg = 0.0048 mol/kg
Now we can substitute these values into the formula for boiling point elevation:
ΔTb = i * Kb * m ΔTb = 3 * 0.51°C/mol * 0.0048 mol/kg = 0.0073°C
So, the predicted change in the boiling point of water when 1.50 g of barium chloride is dissolved in 1.50 kg of water is 0.0073°C. The correct answer is C.0.0073°C.
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