Answer :
To find the predicted change in the boiling point of water when 4.00 grams of barium chloride (BaCl₂) is dissolved in 2.00 kg of water, we need to follow a few steps:
1. Calculate the number of moles of BaCl₂:
[tex]\[ \text{Moles of BaCl}_2 = \frac{\text{Mass of BaCl}_2}{\text{Molar mass of BaCl}_2} \][/tex]
Given:
[tex]\[ \text{Mass of BaCl}_2 = 4.00 \, \text{g} \][/tex]
[tex]\[ \text{Molar mass of BaCl}_2 = 208.23 \, \text{g/mol} \][/tex]
[tex]\[ \text{Moles of BaCl}_2 = \frac{4.00 \, \text{g}}{208.23 \, \text{g/mol}} \approx 0.01921 \, \text{mol} \][/tex]
2. Calculate the molality of the solution:
[tex]\[ \text{Molality} = \frac{\text{Moles of solute}}{\text{Kilograms of solvent}} \][/tex]
Given:
[tex]\[ \text{Kilograms of solvent (water)} = 2.00 \, \text{kg} \][/tex]
[tex]\[ \text{Molality} = \frac{0.01921 \, \text{mol}}{2.00 \, \text{kg}} \approx 0.00960 \, \text{mol/kg} \][/tex]
3. Calculate the change in boiling point ([tex]\(\Delta T_b\)[/tex]):
[tex]\[ \Delta T_b = i \cdot K_b \cdot \text{Molality} \][/tex]
Given:
[tex]\[ i \, \text{(van't Hoff factor for BaCl}_2\text{)} = 3 \][/tex]
[tex]\[ K_b \, \text{(boiling point elevation constant for water)} = 0.51 \, ^\circ \text{C/mol} \][/tex]
[tex]\[ \Delta T_b = 3 \cdot 0.51 \, ^\circ \text{C/mol} \cdot 0.00960 \, \text{mol/kg} \approx 0.01470 \, ^\circ \text{C} \][/tex]
Thus, the change in the boiling point ([tex]\(\Delta T_b\)[/tex]) is approximately [tex]\(0.015^\circ \text{C}\)[/tex].
The correct answer is:
B. [tex]\(0.015^\circ \text{C}\)[/tex]
1. Calculate the number of moles of BaCl₂:
[tex]\[ \text{Moles of BaCl}_2 = \frac{\text{Mass of BaCl}_2}{\text{Molar mass of BaCl}_2} \][/tex]
Given:
[tex]\[ \text{Mass of BaCl}_2 = 4.00 \, \text{g} \][/tex]
[tex]\[ \text{Molar mass of BaCl}_2 = 208.23 \, \text{g/mol} \][/tex]
[tex]\[ \text{Moles of BaCl}_2 = \frac{4.00 \, \text{g}}{208.23 \, \text{g/mol}} \approx 0.01921 \, \text{mol} \][/tex]
2. Calculate the molality of the solution:
[tex]\[ \text{Molality} = \frac{\text{Moles of solute}}{\text{Kilograms of solvent}} \][/tex]
Given:
[tex]\[ \text{Kilograms of solvent (water)} = 2.00 \, \text{kg} \][/tex]
[tex]\[ \text{Molality} = \frac{0.01921 \, \text{mol}}{2.00 \, \text{kg}} \approx 0.00960 \, \text{mol/kg} \][/tex]
3. Calculate the change in boiling point ([tex]\(\Delta T_b\)[/tex]):
[tex]\[ \Delta T_b = i \cdot K_b \cdot \text{Molality} \][/tex]
Given:
[tex]\[ i \, \text{(van't Hoff factor for BaCl}_2\text{)} = 3 \][/tex]
[tex]\[ K_b \, \text{(boiling point elevation constant for water)} = 0.51 \, ^\circ \text{C/mol} \][/tex]
[tex]\[ \Delta T_b = 3 \cdot 0.51 \, ^\circ \text{C/mol} \cdot 0.00960 \, \text{mol/kg} \approx 0.01470 \, ^\circ \text{C} \][/tex]
Thus, the change in the boiling point ([tex]\(\Delta T_b\)[/tex]) is approximately [tex]\(0.015^\circ \text{C}\)[/tex].
The correct answer is:
B. [tex]\(0.015^\circ \text{C}\)[/tex]
