Question
Vapour pressure of CCl _4 at 25^ is 143 mm Hg. If 0.5 g of a non-volatile solute is dissolved in 100 cm ^3 of CCl _4. Find the vapour pressure of the solution. (Density of CCl _4 = 1.58 g / cm ^3 and molecular weight of solute is 65)
MHT CET
Chemistry
Solutions
2026
MHT CET 2026 (11 April Shift 1)
MHT CET Chemistry Question (2026) — Solution
Options
- A. 141.93 mm
- B. 194.39 mm
- C. 199.34 mm
- D. 143.99 mm
Answer
A. 141.93 mm
Step-by-step solution
Mass of solvent (CCl_4), W_1 = Volume Density = 100 1.58 = 158 g Molar mass of CCl_4, M_1 = 12 + 4(35.5) = 154 g/mol Number of moles of solvent, n_1 = W_1 M_1 = 158 154 1.026 Mass of solute, W_2 = 0.5 g Molar mass of solute, M_2 = 65 g/mol Number of moles of solute, n_2 = W_2 M_2 = 0.5 65 0.0077 According to Raoult's law, the relative lowering of vapour pressure is given by: P^0 - P_s P^0 = n_2 n_1 + n_2 143 - P_s 143 = 0.0077 1.026 + 0.0077 143 - P_s 143 = 0.0077 1.0337 0.00745 143 - P_s = 143 0.00745 1.065 P_s = 143 - 1.065 = 141.935 mm Hg
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