Carbon tetrachloride has a vapour pressure of 100 mm Hg at 2 3 ∘ C . This solvent can dissolve candle wax, which is essentially non-volatile. Although candle wax is a mixture, we can take its molecula

Question

Carbon tetrachloride has a vapour pressure of 100 mm Hg at $2 3^{\circ} C$ . This solvent can dissolve candle wax, which is essentially non-volatile.

Although candle wax is a mixture, we can take its molecular formula to be $C_{22} H_{46}$ (molar mass $311 g mol^{- 1}$ ). What is the vapour pressure (in mm Hg ) at $2 3^{\circ} C$ of a solution prepared by dissolving 6.22 g of wax in 77 g of $CCl_{4}$ (molar mass $154 g mol^{- 1}$ )?(Assume Ideal solution)

TestHub · Accepted Answer

To find the vapor pressure of the solution, we use Raoult's Law.

Given:

Vapor pressure of pure CCl₄ ( $P_{CC l_{4}}^{\circ}$ ) = 100 mm Hg

Mass of wax (solute) = 6.22 g

Molar mass of wax (C₂₂H₄₆) = 311 g/mol

Mass of CCl₄ (solvent) = 77 g

Molar mass of CCl₄ = 154 g/mol

1. Calculate moles of CCl₄:

Moles of CCl₄ = $\frac{77 g}{154 g / m o l} = 0.50 m o l$

2. Calculate moles of wax:

Moles of wax = $\frac{6.22 g}{311 g / m o l} = 0.02 m o l$

3. Calculate the total moles:

Total moles = $0.50 m o l + 0.02 m o l = 0.52 m o l$

4. Calculate the mole fraction of CCl₄ ( $X_{CC l_{4}}$ ):

$X_{CC l_{4}} = \frac{M o l eso f CC l _{4}}{T o t a l m o l es} = \frac{0.50 m o l}{0.52 m o l} \approx 0.9615$

5. Apply Raoult's Law:

The vapor pressure of the solution ( $P_{so l u t i o n}$ ) is given by:

$P_{so l u t i o n} = X_{CC l_{4}} \times P_{CC l_{4}}^{\circ}$

$P_{so l u t i o n} = 0.9615 \times 100 mm H g = 96.15 mm H g$

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