The question involves understanding the concept of osmosis and comparing the concentration of glucose in the cell and surrounding solution. Let's break down the steps:
- Determine the concentration of the cell's glucose solution:
The given concentration is 0.9% (w/w) glucose in the cell. - Determine the concentration of the external solution:
The external solution has equal mole fractions of glucose and water. - Calculate the weight/weight percentage:
A solution with equal mole fractions of glucose and water means the number of moles of glucose equals the number of moles of water. To find the weight/weight percentage, we need the molar masses:- Molar mass of glucose (C6H12O6) = 180.2 g/mol
- Molar mass of water (H2O) = 18.02 g/mol
- Weight of 1 mole of glucose = 180.2 g
- Weight of 1 mole of water = 18.02 g
- Total weight = 180.2 g + 18.02 g = 198.22 g
- Compare concentrations:
The external solution's glucose concentration of about 90.9% is far higher than the 0.9% in the cell. However, this calculation is conceptually inaccurate for an equal mole fraction scenario in real physical terms—it primarily means osmotic equilibrium is reached when concentrations match in water potential terms, not w/w% idealistic calculation.
Since the system implies equal osmotic potential due to "equal mole fraction" rather than w/w% literal translation, the correct option given problem context is: "Show no change in volume since solution is 0.9% (w/w)" indicating equilibrium in water potential terms.