Multiple Choice
What is the freezing point depression of a 0.050 m FeCl3 solution at 25.0°C, assuming complete dissociation and using a Kf value of 1.86°C kg/mol for water?
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14. Solutions
Freezing Point Depression
Multiple Choice
Determine the freezing point of a solution that contains 78.8 g of naphthalene (C10H8, molar mass = 128.16 g/mol) dissolved in 722 mL of benzene (density = 0.877 g/mL). Pure benzene has a melting point of 5.50°C and a freezing point depression constant of 4.90°C/m. What is the new freezing point of the solution?
A
0.00°C
B
3.45°C
C
1.23°C
D
5.50°C
Verified step by step guidance1
Calculate the moles of naphthalene (C10H8) using its mass and molar mass. Use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \).
Determine the mass of benzene using its volume and density. Use the formula: \( \text{mass} = \text{volume} \times \text{density} \).
Convert the mass of benzene to kilograms to use in the molality calculation. Remember that 1 g = 0.001 kg.
Calculate the molality of the solution using the formula: \( \text{molality} = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \).
Use the freezing point depression formula: \( \Delta T_f = i \cdot K_f \cdot m \), where \( i \) is the van't Hoff factor (which is 1 for naphthalene), \( K_f \) is the freezing point depression constant, and \( m \) is the molality. Subtract \( \Delta T_f \) from the pure benzene's freezing point to find the new freezing point.
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