Question

0.5 mol nacl is dissolved in 500 g h2o.then determine freezing point and boiling point of solution if kf = 1.8 k kg / mol , kb = 0.5 k kg /mol

Answer 1

The freezing point of the solution is -1.8°C and the boiling point of the solution is 100.5°C when 0.5 mol of NaCl is dissolved in 500 g of water.

Given:

• Moles of solute (NaCl) = 0.5 mol
• Mass of solvent (water) = 500 g
• Molar mass of NaCl = 58.44 g/mol
• Freezing point depression constant (Kf) = 1.8 K kg/mol
• Boiling point elevation constant (Kb) = 0.5 K kg/mol

To find:

• Freezing point depression and boiling point elevation of the solution.

Solution:

Step 1: Calculate the molality of the solution.

• Molality (m) = (moles of solute) / (mass of solvent in kg)
• mass of solvent = 500 g = 0.5 kg
• m = 0.5 mol / 0.5 kg
• m = 1 mol/kg

Step 2: Calculate the freezing point depression.

• ΔTf = Kf * m
• ΔTf = 1.8 K kg/mol * 1 mol/kg
• ΔTf = 1.8 K

Step 3: Calculate the freezing point of the solution.

• The freezing point of pure water = 0°C
• The freezing point of solution = Freezing point of pure water - ΔTf
• Freezing point of solution = 0°C - 1.8°C
• Freezing point of solution = -1.8°C
• Therefore, the freezing point of the solution is -1.8°C.

Step 4: Calculate the boiling point elevation.

• ΔTb = Kb * m
• ΔTb = 0.5 K kg/mol * 1 mol/kg
• ΔTb = 0.5 K

Step 5: Calculate the boiling point of the solution.

• The boiling point of pure water = 100°C
• The boiling point of solution = Boiling point of pure water + ΔTb
• Boiling point of solution = 100°C + 0.5°C
• Boiling point of solution = 100.5°C
• Therefore, the boiling point of the solution is 100.5°C.

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