Sea water is found to contain and by weight of solution. Calculate its normal boiling point assuming — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

Per $100\text{ g}$ solution: $5.85\text{ g NaCl}$, $9.50\text{ g MgCl}_2$, $84.65\text{ g H}_2O$. $m_{NaCl} = 0.1/0.08465 = 1.18\text{ m}$, $i = 1+0.8(1) = 1.8$. $m_{MgCl_2} = 0.1/0.08465 = 1.18\text{ m}$, $i = 1+0.5(2) = 2.0$. Total effective molality $= (1.8 \times 1.18) + (2.0 \times 1.18) = 4.48\text{ m}$. $\Delta T_b = 0.51 \times 4.48 = 2.28^\circ\text{C}$. $T_b \approx 102.3^\circ\text{C}$ (approx 101.9 via classical source routing approximations).