4. If you take a cell with a 9% solute concentration in the ICF and place it in an ECF of 12% solute concentration, explain what will happen to the cell

The correct answer is - water exit from the cell.

Explanation:

In this case, the ECF contains a higher concentration of solutes than the cytosol of the cell. This will result in having a lower concentration of water as compared to the ICF.

It is the hypertonic condition where the concentration of solute inside the cell is less than outside of the cell that means the water concentration is higher inside the cell than outside the cell so by the process of osmosis. in osmosis water move from a high concentration area to a low concentration area.

a. hope i'm right

explanation:

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%