Question

The osmotic pressure of cacl2 and urea solution ofbthe same concentration at the same temperature and respectively 0.605atm and 0.245atm calculate vant hoff factor of cacl2

Answer 1

Given :

The osmotic pressure of CaCl2, Π1= 0.605 M

The osmotic pressure of urea, Π2 = 0.245 M

To Find :

The van't hoff factor of CaCl2.

Calculation:

- As urea is a non-electrolyte, so it will not dissociate and hence its osmotic pressure is given as:

Π2 = C × R × T

⇒ C × R × T = 0.245     ......(i)

- As CaCl2 is an electrolyte, so it will get dissociated and hence its osmotic pressure is given as:

Π1 = i × C × R × T      

As the concentration and temperature is the same, putting the value from eqn (i), we get:

⇒ 0.605 = i × 0.245

⇒ i ≈ 2.47

- So, the Van't hoff factor of CaCl2 is 2.47.

Answer 2

Answer:

Given: Osmotic pressure of CaCl2 solution = 0.605 atm

Osmotic pressure of urea solution = 0.245 atm

To find: The value of van’t Hoff factor

Formulae: π = MRT, π = iMRT

Calculation: For urea solution

π = MRT

0.245 atm = MRT       ....(i)

For CaCl2 solution

π = iMRT

0.602 atm = iMRT       ....(ii)

From equations (i) and (ii),

0.6050.245=iMRTMRT

∴ i = 2.47

The value of van’t Hoff factor is 2.47.