Question

KBr undergoes 80percent dissociation in its 0.5 M aquous solution.Then what is the osmotic presssure of the solution at 27 degree Celsius

Answer 1

Answer:

The osmotic pressure of the solution is 22.14 atm.

Explanation:

Concentration of the solution = C = 0.5M

KBr is 80% dissociated in solution.

$\alpha = 80\%=0.8$

$i=1+(n-1)\alpha$

$i=1+(2-1)0.8=1.8$

Temperature of the solution = 27 °C = 300 K

Osmotic pressure of the solution will be given by:

$\pi=iCRT$

$\pi=1.8\times 0.5M\times 0.0820 atm L/mol K\times 300 K=22.14 atm$

The osmotic pressure of the solution is 22.14 atm.

Answer 2

KBR ------- K^+ + Br^-.

THEREFORE, n=2

NOW

WE KNOW

ALPHA = i-1/n-1

i.e 80/100=i-1/2-1 [as n=2]

I.e, I=0.8+1=1.8

NOW

π = iCRT = 1.8 * 0.5*0.0821*300. [AS TEMPERATURE =27+273K=300K AND R=0.0821 AND CONCENTRATION GIVEN IS 0.5]

THEREFORE , π = 22.14