Assume a living cell with 0.9\%$0.9\%$ (omega/omega)$(\omega/\omega)$ of glucose solution (aqueous). This cell is immersed in another solution having equal mole fraction of glucose and water. (Consider the data upto first decimal place only) The cell will:

### Related Formula Mass percentage from mole fraction calculation: \%text w/w = fracx_1 cdot M_1x_1 cdot M_1 + x_2 cdot M_2 times 100$$\%\text{ w/w} = \frac{x_1 \cdot M_1}{x_1 \cdot M_1 + x_2 \cdot M_2} \times 100$$ ### Core Logic Inside the living cell, glucose concentration is 0.9\%text w/w$0.9\%\text{ w/w}$. The surrounding solution has equal mole fractions of glucose and water (x_textglucose = 0.5$x_{\text{glucose}} = 0.5$, x_textwater = 0.5$x_{\text{water}} = 0.5$). Let's calculate the mass percentage of the outer solution: - Mass of glucose component = 0.5 times 180 = 90mathrm\ g$0.5 \times 180 = 90\mathrm{\ g}$ - Mass of water component = 0.5 times 18 = 9mathrm\ g$0.5 \times 18 = 9\mathrm{\ g}$ - Total solution mass = 90 + 9 = 99mathrm\ g$90 + 9 = 99\mathrm{\ g}$ ### Step 1: Concentration Determination and Osmosis Profile Outer mass percentage: \%text w/w = frac9099 times 100 approx 90.9\%$$\%\text{ w/w} = \frac{90}{99} \times 100 \approx 90.9\%$$ Because the external environment is highly concentrated (hypertonic) compared to the inner cell (0.9\%$0.9\%$), water flows out of the cell via exosmosis, causing the **cell to shrink**. Note: Because the reasoning establishes shrinkage but the quantitative figures in the options are highly mismatched, this question is officially designated as a **Bonus** question. ### Pattern Recognition An equal mole fraction solution of a high-molar-mass solute (glucose, 180mathrm\ g/mol$180\mathrm{\ g/mol}$) and a low-molar-mass solvent (water, 18mathrm\ g/mol$18\mathrm{\ g/mol}$) is always extremely concentrated by mass. Placing a standard living cell into such a hypertonic solution inevitably causes fluid loss and cellular shrinkage. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions