A ‘100 proof’ solution of ethanol in water consists of 50.00 ml of and 50.00 ml of mixed at . The de — Solutions and Colligative Properties Chemistry Question
Question
A ‘100 proof’ solution of ethanol in water consists of 50.00 ml of $\text{C}_2\text{H}_5\text{OH(l)}$ and 50.00 ml of $\text{H}_2\text{O(l)}$ mixed at $15.56^\circ\text{C}$. The density of the solution is $0.9344 \text{ g/ml}$, that of pure $\text{H}_2\text{O}$ is $1.0000 \text{ g/ml}$ and that of pure $\text{C}_2\text{H}_5\text{OH}$ is $0.7939 \text{ g/ml}$. Is the solution ideal? Answer ‘1’, if the solution is ideal and answer ‘2’, if the solution is non-ideal.
💡 Solution & Explanation
Mass of ethanol = $50.00 \text{ ml} \times 0.7939 \text{ g/ml} = 39.695 \text{ g}$. Mass of water = $50.00 \text{ ml} \times 1.0000 \text{ g/ml} = 50.00 \text{ g}$. Total mass of solution = $39.695 + 50.00 = 89.695 \text{ g}$. The actual volume of the solution is $V = \text{Mass} / \text{Density} = 89.695 \text{ g} / 0.9344 \text{ g/ml} \approx 96.0 \text{ ml}$. The expected ideal volume was $50.00 + 50.00 = 100.00 \text{ ml}$. Since $V_{actual} < V_{ideal}$, there is a volume contraction ($\Delta V_{mix} < 0$), indicating the solution is non-ideal (negative deviation). Therefore, correct answer is 2.