The densities of ice and water at and 1 bar are and , respectively. If the percentage of occupied sp — Solid State Chemistry Question

💡 Solution & Explanation

Let mass be $M$. $V_{ice} = M/0.96$. $V_{water} = M/0.99$. Occupied volume $V_{occ} = (x/100) \times V_{ice} = (x/100) \times (M/0.96)$. Fraction of occupied space in water = $V_{occ} / V_{water} = \frac{(x/100)(M/0.96)}{M/0.99} = (x/100) \times (0.99/0.96) = \frac{33x}{3200}$. Percentage of occupied space in water = $\frac{33}{32}x$. Percentage of empty space = $100 - \frac{33}{32}x$. Therefore, correct answer is C.