Question

Calculate osmotic pressure of CaCl 2 solution at 36ºC, if its molarity is 0.02
mol/L, degree of dissociation is 68 %.​

Answer 1

we have to calculate osmotic pressure of CaCl₂ solution at 36°C , it its molarity is 0.02 mol/L and degree of dissociation is 68%.

solution : dissociation reaction of CaCl₂ is ..

CaCl₂ ⇔ Ca²⁺ + 2Cl¯

1 - α α 2α

so, i = 1 -α + α + 2α = 1 + 2α

here degree of dissociation, α = 68% = 0.68

so, i = 1 + 2 × 0.68 = 1 + 1.36 = 2.36

now using formula,

osmotic pressure, π = iMRT

here i = 2.36, M = 0.02 mol/L and T = 36°C = 273 + 36 = 309K

now π = 2.36 × 0.02 × 0.082 × 309 atm

= 1.1959 ≈ 1.2 atm

Therefore the osmotic pressure would be 1.2 atm