Question

Part #1:

Imagine a red blood cell with a low concentration (~0.9%) of solutes dissolved in the water in the cell. The cell membrane acts as a semi-permeable membrane. If this cell is placed in brine (a highly concentrated solution of sodium chloride) which describes what will occur?

  A Nothing will occur

B Sodium chloride will flow from the cell into the brine

C Water will flow from the brine into the cell

D Sodium chloride will flow from the brine into the cell

E Water will flow from the cell into the brine

Part #2:

An isotonic solution of glucose contains 4.82 g of glucose (C6H12O6, Molar Mass = 180.16 g/mol) in 100.0 mL of solution. Imagine this solution is separated from pure water by a semi-permeable membrane. What would the osmotic pressure of the solution be at 297.0 K? Enter your answer in the unit of atmospheres.

Answer 1

Answer Part 1)

Option E is correct: Water will flow from cell to brine.

Because cell membrane is semi permeable and only solvent can flow through it. And since the concentration of water inside is high (outside solute concentration is high) water will flow from cell to brine till the concentration of water on both side is equal.

Answer Part B)

The osmosis equation is:

π = iMRT

π is not equal to 3.14159 in this situation, π stands for the osmotic pressure and is usually expressed in the pressure unit of atmospheres.

The lowercase letter "i" is called the van 't Hoff factor

M is molarity

R is the gas constant and we will be using the same value as in the gas laws unit: 0.08206 L atm/mol K

for non-ionic compounds in solution, like glucose (C6H12O6) , the van't Hoff factor is 1, i.e i= 1

In the question ,

180.16 g in 1000 ml = 1M

4.82 g in 100 ml = ? M

4.82 g in 100 ml = 0.26 M = 0.26 mol/L

π = 1 × (0.26 mol/L) (0.08206 L atm / mol K) (297 K)

π = 6.413 atm

osmotic pressure of the solution be at 297.0 K = π = 6.413 atm