Vapour Pressure of CCl4 Solution

Given:
Temperature (T) = 25°C
Vapour pressure of pure CCl4 (P°) = 143 mm Hg
Mass of solute (m) = 0.5 g
Molar mass of solute (M) = 65 g/mol
Volume of CCl4 (V) = 100 mL
Density of CCl4 = 1.58 g/cm³

To find:
Vapour pressure of the solution (P solution)

We can use Raoult's Law to calculate the vapour pressure of the solution.

Raoult's Law states that the partial pressure of each component in an ideal mixture of liquids is equal to the product of the vapor pressure of the pure component and its mole fraction in the mixture.

Mathematically:
P solution = P° solvent × X solvent

Where:
P solution is the vapour pressure of the solution
P° solvent is the vapour pressure of the pure solvent
X solvent is the mole fraction of the solvent in the solution

Now, let's calculate the mole fraction of the solvent (CCl4) in the solution.

1. Calculate the moles of CCl4:
Moles of CCl4 = mass of CCl4 / molar mass of CCl4
= 100 mL × 1.58 g/cm³ / 153.82 g/mol
= 1.026 mol

2. Calculate the moles of solute (non-volatile):
Moles of solute = mass of solute / molar mass of solute
= 0.5 g / 65 g/mol
= 0.0077 mol

3. Calculate the mole fraction of CCl4:
Mole fraction of CCl4 = moles of CCl4 / (moles of CCl4 + moles of solute)
= 1.026 mol / (1.026 mol + 0.0077 mol)
= 0.9925

Now, substitute the values into Raoult's Law equation:

P solution = P° solvent × X solvent
= 143 mm Hg × 0.9925
= 142.5 mm Hg

Therefore, the vapour pressure of the solution is 142.5 mm Hg.