What is the osmotic pressure of a 0.201 M solution of MgCl2 at 37.0 °C? (assume complete dissociation).

Question

Question

Answer 1

Answer #1

Molarity of the solution (M) = 0.201 M = 00.201 mol.L^–1

MgCl2 dissociates into Mg⁺² and 2 Cl^– ( total 3 ions)

So, van't Hoff factor is i = 3

The osmotic pressure = π = iMRT

where M= molarity = 0.201 M

R = 0.0821 L.atm.mol^-1.K^-1.

T =37.0 °C = 37.0 + 273.15 = 310.15 K

Osmotic pressure (π) =3 × 0.201 mol.L^-1.× 0.0821 L.atm.mol^-1.K^-1.× 310.15 K

= 15.35438 atm

Osmotic pressure of the solution = 15.4 atm

Answer 2

Answer #2

Answer 3

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