Given:
mass of solute = 10.3 g
mass of solvent = 250 g = 0.250 kg
solute: glucose (C₆H₁₂O₆)
solvent: water (H₂O)
Required:
[tex]\Delta T_{\text{f}}[/tex]
Solution:
Step 1: Calculate the molar mass of solute.
molar mass of solute = (12.01 g/mol × 6) + (1.008 g/mol × 12) + (16.00 g/mol × 6)
molar mass of solute = 180.156 g/mol
Step 2: Calculate the number of moles of solute.
[tex]\text{moles of solute} = \frac{\text{mass of solute}}{\text{molar mass of solute}}[/tex]
[tex]\text{moles of solute} = \frac{\text{10.3 g}}{\text{180.156 g/mol}}[/tex]
[tex]\text{moles of solute} = \text{0.0571726726 mol}[/tex]
Step 3: Calculate the molality of solution.
[tex]m = \frac{\text{moles of solute}}{\text{mass of solvent}}[/tex]
[tex]m = \frac{\text{0.0571726726 mol}}{\text{0.250 kg}}[/tex]
[tex]m = \text{0.22869069 molal}[/tex]
Step 4: Calculate the freezing point depression.
Note: The value of Kf for water is 1.86°C/molal.
[tex]\Delta T_{\text{f}} = mK_{\text{f}}[/tex]
[tex]\Delta T_{\text{f}} = (\text{0.22869069 molal})(\text{1.86°C/molal})[/tex]
[tex]\boxed{\Delta T_{\text{f}} = \text{0.425°C}}[/tex]
[tex]\\[/tex]
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