At 100°C, the vapour pressure of a solution of 4.0 g of solute in 100 g of water is 750 mm. Boiling — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

Relative lowering of vapour pressure at 100°C (where $P^\circ = 760$ mm) is $(760 - 750) / 750 = n / N$. Moles of water $N = 100 / 18 = 5.555$ mol. $10 / 750 = n / 5.555 \Rightarrow n = 5.555 / 75 = 0.07407$ mol. Molality $m = n / 0.1 \text{ kg} = 0.7407$ m. Boiling point elevation $\Delta T_b = K_b \times m = 0.52 \times 0.7407 \approx 0.385^\circ C \approx 0.4^\circ C$. Thus, boiling point is $100 + 0.4 = 100.4^\circ C$. Therefore, correct answer is C.