Equal masses of two aqueous solutions, one of 3.6 g glucose per 100 g of water and other of 3.6 g ur — Solutions and Colligative Properties Chemistry Question
Question
Equal masses of two aqueous solutions, one of 3.6 g glucose per 100 g of water and other of 3.6 g urea per 100 g of water are mixed. If the freezing point of resulting solution is $-x^\circ\text{C}$, then the value of ‘1000$x$’ is ($K_f \text{ of water} = 1.86$)
💡 Solution & Explanation
Solution 1: 3.6 g glucose + 100 g water = 103.6 g. Solution 2: 3.6 g urea + 100 g water = 103.6 g. Mixing equal masses (say 103.6 g of each) gives a total mass of 207.2 g containing 200 g (0.2 kg) of water. Moles of glucose = $3.6 / 180 = 0.02 \text{ mol}$. Moles of urea = $3.6 / 60 = 0.06 \text{ mol}$. Total moles of solute = $0.02 + 0.06 = 0.08 \text{ mol}$. Final molality $m = 0.08 \text{ mol} / 0.2 \text{ kg} = 0.4 \text{ m}$. $\Delta T_f = K_f \times m = 1.86 \times 0.4 = 0.744^\circ\text{C}$. The freezing point is $-0.744^\circ\text{C}$, so $x = 0.744$. The value of $1000x = 744$. Therefore, correct answer is 0744.