To calculate the molal elevation constant (\(K_b\)) for water and the boiling point elevation for a 0.1 molal urea solution, we can use the formula:
\[
K_b = \dfrac{R \cdot T_b^2}{\Delta H_v}
\]
Where:
- \(R\) is the gas constant, \(R = 1.987 \dfrac{cal}{mol \cdot K}\)
- \(T_b\) is the boiling point of water in Kelvin, \(T_b = 373.15 K\)
- \(\Delta H_v\) is the latent heat of vaporization of water, \( \Delta H_v = 9.72 \dfrac{kcal}{mol} = 9720 \dfrac{cal}{mol}\)
### Step 1: Calculate the molal elevation constant \(K_b\)
Substitute the values into the equation:
\[
K_b = \dfrac{1.987 \cdot (373.15)^2}{9720}
\]
Let's calculate this first.
The molal elevation constant, \(K_b\), is approximately \(0.512 \, \text{kg mol}^{-1} \, \text{K}^{-1}\).
### Step 2: Boiling Point Elevation
The boiling point elevation can be calculated using:
\[
\Delta T_b = K_b \cdot m
\]
Where:
- \(K_b = 0.512 \, \text{kg mol}^{-1} \, \text{K}^{-1}\)
- \(m = 0.1 \, \text{molal}\)
Now, let's calculate the boiling point elevation \(\Delta T_b\).
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However, you can manually calculate the boiling point elevation using the formula:
\[
\Delta T_b = K_b \cdot m
\]
Substituting the values:
\[
\Delta T_b = 0.512 \, \text{K/molal} \times 0.1 \, \text{molal} = 0.0512 \, \text{K}
\]
So, the new boiling point of the solution is:
\[
T_b = 373.15 \, \text{K} + 0.0512 \, \text{K} = 373.20 \, \text{K}
\]
The correct answer is:
**(A)** \( K_b = 0.512 \, \text{kg mol}^{-1} \, \text{K}^{-1}, T_b = 373.20 \, \text{K} \).
