calculate de osmotic peessure of 7.40 mM MgCl2

Question

Question

science chemistry

Answer 1

Answer #1

Sol.

As The Conc. of MgCl2 = C = 7.40 mM = 7.40 × 10^-3 M

Dissociation reaction of MgCl2 :

MgCl2 ----> Mg2+ + 2Cl-

So , Van't Hoff factor = i = Total Number of ions produced

= 1 + 2 = 3

Gas constant = R = 0.0821 L atm / K mol

Temeprature = T

So , Osmotic Pressure = P = iCRT

= 3 × 7.40 × 10^-3 × 0.0821 × T

= 0.00182262 × T

As the temperature , T( in K ) is not given in the question , so , i am writing it as only T

Answer 2

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