Assume a living cell with 0.9 % (omega / omeg | vedclass

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(B) The living cell contains $0.9 \% (\omega / \omega)$ glucose solution. For the external solution,the mole fraction of glucose $(x_g)$ and water $(x

Vedclass · Accepted Answer

shrink since solution is $0.45 \% (\omega / \omega)$ as a result of association of glucose molecules (due to hydrogen bonding)

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(B) The living cell contains $0.9 \% (\omega / \omega)$ glucose solution. For the external solution,the mole fraction of glucose $(x_g)$ and water $(x_w)$ are equal,so $x_g = x_w = 0.5$. Let the number of moles of glucose be $0.5$ and water be $0.5$. Mass of glucose $= 0.5 \ mol 	imes 180 \ g/mol = 90 \ g$. Mass of water $= 0.5 \ mol 	imes 18 \ g/mol = 9 \ g$. Total mass of solution $= 90 \ g + 9 \ g = 99 \ g$. Percentage by mass $(\omega / \omega) = (	ext{mass of solute} / 	ext{total mass of solution}) 	imes 100 = (90 / 99) 	imes 100 \approx 90.9 \%$. Since the external solution $(90.9 \%)$ is hypertonic compared to the cell $(0.9 \%)$,water will move out of the cell due to osmosis. Therefore,the cell will shrink.

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

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