## Related questions with answers

Plants that thrive in salt water must have internal solutions (inside the plant cells) that are isotonic with (have the same osmotic pressure as) the surrounding solution. A leaf of a saltwater plant is able to thrive in an aqueous salt solution (at $\left.25^{\circ} \mathrm{C}\right)$ that has a freezing point equal to $-0.621^{\circ} \mathrm{C} .$ You would like to use this information to calculate the osmotic pressure of the solution in the cell. Solve for the osmotic pressure (at $\left.25^{\circ} \mathrm{C}\right)$ of the solution in the plant cell.

Solutions

VerifiedThis exercise revolves around **osmotic pressure** of the solution inside the plant cells.

In the textbook, it was stated that we have a solution that has a freezing point of -0.621$\degree$C. Also, it was stated that at 25$\degree$C erythrocytes neither swell nor shrivel in this solution.

We must note that plant cells have **the same osmotic pressure** as the surrounding solution (isotonic solution).

Also, to calculate the osmotic pressure, we need to assume that the molarity and molality for this instance are the same.

*Do you recall what are and how to calculate the freezing-point depression and osmotic pressure?*

Freezing point depression ($\triangle T$) = 0.621

$K_{f}$ for water = 1.86

from formula --

$\triangle T$ = $K_{f} m_{solute}$

$m_{solute}$ = $\frac{0.621}{1.86}$ = 0.33

.

According to assumption

M = m = 0.33

.

Now osmatic pressure at temperature 25$\text{\textdegree}$C -

$\Pi$ = MRT

$\Pi$ = $0.33\times0.083\times298$ = 8.16 atm

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