1. A solution contains of 10.3 g of glucose (C6 H12O6 ) dissolved in 250 g of water., What is the freezing point depression of the solvent?

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1. A solution contains of 10.3 g of glucose (C6 H12O6 ) dissolved in 250 g of water.


What is the freezing point depression of the solvent?​

Given:

mass of solute = 10.3 g

mass of solvent = 250 g = 0.250 kg

solute: glucose (C₆H₁₂O₆)

solvent: water (H₂O)

Required:

\Delta T_{\text{f}}

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.

\text{moles of solute} = \frac{\text{mass of solute}}{\text{molar mass of solute}}

\text{moles of solute} = \frac{\text{10.3 g}}{\text{180.156 g/mol}}

\text{moles of solute} = \text{0.0571726726 mol}

Step 3: Calculate the molality of solution.

m = \frac{\text{moles of solute}}{\text{mass of solvent}}

m = \frac{\text{0.0571726726 mol}}{\text{0.250 kg}}

m = \text{0.22869069 molal}

Step 4: Calculate the freezing point depression.

Note: The value of Kf for water is 1.86°C/molal.

\Delta T_{\text{f}} = mK_{\text{f}}

\Delta T_{\text{f}} = (\text{0.22869069 molal})(\text{1.86°C/molal})

\boxed{\Delta T_{\text{f}} = \text{0.425°C}}

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