To determine the predicted change in the boiling point of water when 4.00 g of barium chloride ([tex]\( \text{BaCl}_2 \)[/tex]) is dissolved in 2.00 kg of water, we can follow a series of calculations:
1. Calculate the number of moles of [tex]\( \text{BaCl}_2 \)[/tex] dissolved:
[tex]\[
\text{moles of } \text{BaCl}_2 = \frac{\text{mass of } \text{BaCl}_2}{\text{molar mass of } \text{BaCl}_2}
\][/tex]
Given:
[tex]\[
\text{mass of } \text{BaCl}_2 = 4.00 \, \text{g}
\][/tex]
[tex]\[
\text{molar mass of } \text{BaCl}_2 = 208.23 \, \text{g/mol}
\][/tex]
The number of moles of [tex]\( \text{BaCl}_2 \)[/tex] is:
[tex]\[
\text{moles of } \text{BaCl}_2 = \frac{4.00 \, \text{g}}{208.23 \, \text{g/mol}} \approx 0.0192 \, \text{moles}
\][/tex]
2. Calculate the molality of the solution:
[tex]\[
\text{molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}
\][/tex]
Given:
[tex]\[
\text{mass of water} = 2.00 \, \text{kg}
\][/tex]
The molality of the solution is:
[tex]\[
\text{molality} = \frac{0.0192 \, \text{moles}}{2.00 \, \text{kg}} \approx 0.0096 \, \text{mol/kg}
\][/tex]
3. Calculate the change in boiling point using the boiling point elevation formula:
[tex]\[
\Delta T_b = i \times K_b \times m
\][/tex]
Given:
[tex]\[
i (\text{van't Hoff factor for } \text{BaCl}_2) = 3
\][/tex]
[tex]\[
K_b (\text{boiling point elevation constant}) = 0.51 \, ^\circ\text{C/mol}
\][/tex]
[tex]\[
m (\text{molality}) = 0.0096 \, \text{mol/kg}
\][/tex]
The change in boiling point is:
[tex]\[
\Delta T_b = 3 \times 0.51 \, ^\circ\text{C/mol} \times 0.0096 \, \text{mol/kg} \approx 0.015 \, ^\circ\text{C}
\][/tex]
Thus, the predicted change in the boiling point of water is approximately [tex]\(0.015^\circ C\)[/tex].
So, the correct answer is:
D. [tex]\(0.015^\circ C\)[/tex]
