When cells of the skeletal vacuole of a frog were placed in a series of NaCl solutions of different — Solutions and Colligative Properties Chemistry Question
Question
When cells of the skeletal vacuole of a frog were placed in a series of NaCl solutions of different concentrations at $6^\circ\text{C}$, it was observed microscopically that they remained unchanged in $x\%$ NaCl solution, shrank in more concentrated solutions and swells in more dilute solutions. Water freezes from the $x\%$ salt solution at $-0.40^\circ\text{C}$. If the osmotic pressure of the cell cytoplasm at $6^\circ\text{C}$ is ‘$y$’ $\times$ 0.0821 atm, then the value of ‘$y$’ is ($K_f = 1.86 \text{ K mol}^{-1}\text{kg}$)
💡 Solution & Explanation
The cell fluid is isotonic with the $x\%$ NaCl solution, meaning they have the same effective concentration and osmotic pressure. The freezing point of this solution is $-0.40^\circ\text{C}$, so $\Delta T_f = 0.40^\circ\text{C}$. Using $\Delta T_f = K_f \times m_{eff} \implies 0.40 = 1.86 \times m_{eff} \implies m_{eff} = 0.40 / 1.86 \approx 0.215 \text{ mol/kg}$. For dilute aqueous solutions, molarity $M_{eff} \approx m_{eff} = 0.215 \text{ M}$. The osmotic pressure at $6^\circ\text{C}$ (279 K) is $\pi = M_{eff} R T = 0.215 \times R \times 279 = 60.0 \times R$. Given $\pi = y \times 0.0821$ (where $0.0821$ is $R$), we have $y = 60$. Therefore, correct answer is 0060.