Question

Vapour pressure of \(\text{CCl}_4\) at \(25^{\circ}\)C is \(143\) mm Hg. If \(0.5\) g of a non-volatile solute is dissolved in \(100\text{ cm}^3\) of \(\text{CCl}_4\). Find the vapour pressure of the solution.
(Density of \(\text{CCl}_4 = 1.58\text{ g}/\text{cm}^3\) and molecular weight of solute is \(65\))

• A. \(141.93\) mm
• B. \(194.39\) mm
• C. \(199.34\) mm
• D. \(143.99\) mm

Hint:

Use Raoult's law: $\frac{P^0-P}{P^0}=x_{\text{solute}}$.

Answer 1

The Correct Option is A

Step 1: Find the moles
Mass of CCl\(_4\) \(=100\times1.58=158\) g. Molar mass of CCl\(_4\) \(=154\), so \(n_{\text{solvent}}=\frac{158}{154}=1.026\) mol.
Moles of solute \(=\frac{0.5}{65}=0.00769\) mol.

Step 2: Mole fraction of solute
\[ x_2=\frac{0.00769}{0.00769+1.026}=0.00744 \]

Step 3: Lowering of vapour pressure
\[ P^0-P=0.00744\times143=1.064\text{ mm} \]
\[ P=143-1.064=141.93\text{ mm} \]

Step 4: Check
Other options exceed \(143\) mm, which is impossible since a solute lowers the pressure. Option (A).

Final Answer:
The vapour pressure is \(141.93\) mm Hg, option (A). \[ \boxed{\text{(A)}} \]