Sure! Let's go through this step-by-step:
### Given Data:
1. Weight of acetone ([tex]\( W_1 \)[/tex]) = 26.68 g
2. Weight of camphor ([tex]\( W_2 \)[/tex]) = 1.21 g
3. Molar mass of camphor = 152 g/mol
4. Boiling point of the solution = 329.95 K
5. Boiling point of pure acetone = 329.45 K
### Steps to Solve:
1. Calculate the Change in Boiling Point ([tex]\( \Delta T_b \)[/tex])
[tex]\[
\Delta T_b = \text{Boiling point of solution} - \text{Boiling point of pure acetone}
\][/tex]
[tex]\[
\Delta T_b = 329.95 \, \text{K} - 329.45 \, \text{K} = 0.50 \, \text{K}
\][/tex]
The change in boiling point [tex]\( \Delta T_b \)[/tex] is 0.50 K.
2. Calculate the Number of Moles of Camphor ([tex]\( n_2 \)[/tex])
[tex]\[
n_2 = \frac{\text{Weight of camphor} (W_2)}{\text{Molar Mass of camphor}}
\][/tex]
[tex]\[
n_2 = \frac{1.21 \, \text{g}}{152 \, \text{g/mol}} = 0.00796053 \, \text{mol}
\][/tex]
The number of moles of camphor [tex]\( n_2 \)[/tex] is approximately 0.00796053 mol.
3. Convert the Mass of Solvent (Acetone) to Kilograms ([tex]\( m_{\text{solvent}} \)[/tex])
[tex]\[
m_{\text{solvent}} = \frac{\text{Weight of acetone} (W_1)}{1000}
\][/tex]
[tex]\[
m_{\text{solvent}} = \frac{26.68 \, \text{g}}{1000} = 0.02668 \, \text{kg}
\][/tex]
The mass of solvent in kilograms is 0.02668 kg.
4. Calculate the Molality ([tex]\( m \)[/tex]) of the Solution
[tex]\[
m = \frac{n_2}{m_{\text{solvent}}}
\][/tex]
[tex]\[
m = \frac{0.00796053 \, \text{mol}}{0.02668 \, \text{kg}} = 0.29837 \, \text{mol/kg}
\][/tex]
The molality of the solution is approximately 0.29837 mol/kg.
5. Calculate the Molal Elevation Constant ([tex]\( K_b \)[/tex])
[tex]\[
K_b = \frac{\Delta T_b}{m}
\][/tex]
[tex]\[
K_b = \frac{0.50 \, \text{K}}{0.29837 \, \text{mol/kg}} = 1.67577 \, \text{K} \cdot \text{kg/mol}
\][/tex]
The molal elevation constant [tex]\( K_b \)[/tex] for acetone is approximately 1.67577 K·kg/mol.
Therefore, the molal elevation constant ([tex]\( K_b \)[/tex]) for acetone is 1.67577 K·kg/mol.
