Question

Osmotic pressure is the external pressure which should be applied to stop the flow of solvent into the solution when the two are separated by a semipermeable membrane. The osmotic pressure is a colligative property. Two solutions having the same osmotic pressure are called isotonic. If there are two solutions and one of them is of lower osmotic pressure, it is called hypotonic while the other is called hypertonic.

Answer the questions below:

1. What will happen if red blood corpuscles are placed in a 5% NaCl solution which is a hypertonic solution?    [1]
2. Show that osmotic pressure (π) is a colligative property.    [1]
3. Calculate the amount of pressure required to stop osmosis of a solution when 40 g of Na₂SO₄ is added to 1 L of water at 298 K.
   (Given: Na = 23, O = 16, S = 32 and R = 0.0821 L atm K⁻¹ mol⁻¹)    [2]
4. Briefly discuss the process of reverse osmosis followed to desalinate sea water and convert it into drinking water.    [1]

Answer 1

a. As compared to the cytoplasm of red blood cells, a hypertonic fluid has a higher osmotic pressure (higher solute concentration). Osmosis causes water to flow from an area of lower solute concentration (within the RBCs) to a region of higher solute concentration (the surrounding NaCl solution) when red blood cells (RBCs) are submerged in a 5% NaCl solution, which is hypertonic. RBCs shrink as a result, a process known as crenation.

b. Colligative property depends only on the number of solute particles and not on their nature. Osmotic pressure (π) is given by Van’t Hoff’s equation as follows:

π = iCRT

Where,

i = Van’t Hoff factor (degree of dissociation/association)

C = Concentration (molarity) of solute

R = Universal gas constant

T = Absolute temperature in Kelvin

Osmotic pressure is a colligative feature since it is dependent on concentration (number of solute particles per unit volume).

c. Given: Mass of Na₂SO₄ = 40 g

Now, first we will calculate the molar mass of Na₂SO₄,

Molar mass of Na₂SO₄:

M = (2 × 23) + (1 × 32) + (4 × 16)

= 46 + 32 + 64

= 142 g/mol

Now, moles of Na₂SO₄ will be:

= `"Mass"/"Molar mass"`

= `40/142`

= 0.2817 moles

Now, we will determine the Van’t Hoff factor (i),

Na₂SO₄ dissociates completely in water as:

\[\ce{Na2SO4 -> 2Na+ + SO^{2-}_4}\]

This produces 3 particles per molecule, so i = 3.

Now, we will calculate the osmotic pressure,

Using Van’t Hoff’s equation:

π = iCRT

= (3) × (0.2817) × (0.0821) × (298)

= 3 × 0.2817 × 24.5058

= 20.7 atm

Thus, the required pressure to stop osmosis is 20.7 atm.

d. Reverse osmosis (RO) is a desalination method that uses high pressure to force water through a semipermeable membrane in order to remove salts and other impurities from seawater. Water typically transitions from low solute concentration (pure water) to high solute concentration (seawater) by osmosis. In reverse osmosis, salts and contaminants are left behind as water molecules are forced to travel from a high solute concentration (seawater) to a lower solute concentration (clean water) under an external pressure higher than the osmotic pressure. Fresh drinking water is produced as a result, while the concentrated brine is discarded. Reverse osmosis is a common method used in industries and at home to purify water.