The solution of a non-volatile solute in water freezes at −0.30°C. The vapour pressure of pure water is 23.51 mm Hg and Kf for water is 1.86 degrees/molal. What is the vapour pressure of the solution at 298 K?
Answer & explanation
Correct answer: option 3
The correct answer is option 3. 23.442 mm
Given, Vapour pressure of the pure solvent, \(p_0 = 23.1 mm\)
Cryoscopic constant, \(K_f = 1.86\) degree/molal
Depression in freezing point, \(\Delta T = −0.30^oC\)
\(\Delta T = K_f\text{ × }Molality\)
⇒ \(0.3 = 1.86\text{ × }Molality\)
∴ \(\text{Molality = }0.161\)
Now,
\(Molality \text{ = }\frac{X_A × 1000}{(1 − X_A) × m_B}\)
⇒\(0.161 \text{ = }\frac{X_A × 1000}{(1 − X_A) × 18}\)
⇒\(1 − X_A \text{ = }\frac{X_A × 1000}{0.161 × 18}\)
⇒\(1 − X_A \text{ = }\frac{X_A × 1000}{2.898}\)
⇒\(1 − X_A \text{ = }X_A × 345.06\)
⇒\(\frac{1}{X_A} − 1 = 345.06\)
⇒\(\frac{1}{X_A} = 346.06\)
⇒\(X_A = \frac{1}{346.06}\)
∴ \(X_A = 0.00289\)
Also,
\(\frac{\Delta p}{p_0} = X_A\)
⇒\(\frac{p_0 − p_s}{p_0} = X_A\)
⇒\(\frac{23.51 − p_s}{23.51} = 0.00289\)
⇒\(\frac{23.51 − p_s}{23.51} = 0.00289\)
⇒\(23.51 − p_s = 0.00289 × 23.51\)
⇒\(23.51 − p_s = 0.00289 × 23.51\)
⇒\(p_s = 23.51 − (0.00289 × 23.51)\)
∴ \(p_s = 23.442 mm\)