Calculate the relative lowering of vapour pressure of a 10% aqeous caustic soda solution.

Answer

To calculate the relative lowering of vapor pressure of a 10% aqueous caustic soda (NaOH) solution, we can follow these steps:

### Step 1: Determine the mass of solute and solvent
In a 10% aqueous solution, we have:
- Mass of NaOH (solute) = 10 g
- Mass of water (solvent) = 90 g

### Step 2: Calculate the molar mass of NaOH and water
- Molar mass of NaOH = Na (23) + O (16) + H (1) = 40 g/mol
- Molar mass of water (H₂O) = H (1) × 2 + O (16) = 18 g/mol

### Step 3: Calculate the number of moles of NaOH and water
- Number of moles of NaOH = mass / molar mass = 10 g / 40 g/mol = 0.25 moles
- Number of moles of water = mass / molar mass = 90 g / 18 g/mol = 5 moles

### Step 4: Calculate the total number of moles in the solution
Total moles = moles of NaOH + moles of water
Total moles = 0.25 + 5 = 5.25 moles

### Step 5: Calculate the mole fraction of NaOH
Mole fraction of NaOH = moles of NaOH / total moles
Mole fraction of NaOH = 0.25 / 5.25 ≈ 0.0476

### Step 6: Determine the relative lowering of vapor pressure
The relative lowering of vapor pressure is equal to the mole fraction of the solute (NaOH).
Relative lowering of vapor pressure = 0.0476 (approximately 0.048)

### Final Answer
The relative lowering of vapor pressure of a 10% aqueous caustic soda solution is approximately 0.048 (unitless).

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