An aqueous solution contains 12% (w/w) and 9.5% (w/w) . If the sulphate dissociates up to 80% and ch — Solutions and Colligative Properties Chemistry Question
Question
An aqueous solution contains 12% (w/w) $\text{MgSO}_4$ and 9.5% (w/w) $\text{MgCl}_2$. If the sulphate dissociates up to 80% and chloride dissociates up to 60%, the boiling point of solution (in Kelvin) is ($K_b \text{ of water} = 0.785 \text{ K-kg/mol}$)
💡 Solution & Explanation
Consider 100 g of solution: 12 g $\text{MgSO}_4$ ($M=120$), 9.5 g $\text{MgCl}_2$ ($M=95$), and 78.5 g water. Moles $\text{MgSO}_4 = 12/120 = 0.1 \text{ mol}$. Moles $\text{MgCl}_2 = 9.5/95 = 0.1 \text{ mol}$. For $\text{MgSO}_4$ ($n=2, \alpha=0.8$), $i_1 = 1 + 0.8 = 1.8$. For $\text{MgCl}_2$ ($n=3, \alpha=0.6$), $i_2 = 1 + 2(0.6) = 2.2$. Total effective moles = $(0.1 \times 1.8) + (0.1 \times 2.2) = 0.18 + 0.22 = 0.40 \text{ mol}$. Mass of water = 78.5 g = 0.0785 kg. Molality $m = 0.40 / 0.0785 = 5.0955 \text{ m}$. $\Delta T_b = K_b \times m = 0.785 \times 5.0955 = 4.00^\text{K}$. Boiling point $= 100^\circ\text{C} + 4^\circ\text{C} = 104^\circ\text{C}$. In Kelvin, $T = 104 + 273.15 = 377.15 \text{ K}$. Rounding to integer gives 377. Therefore, correct answer is 0377.