Question

Calculate the amount of CaCl₂ (van't Hoff factor i = 2.47) dissolved in 2.5 L solution so that its osmotic pressure at 300K is 0.75 atmosphere.

Given : Molar mass of CaCl₂ is 111g mol^-1

R=0.082 L.atm K^-1mol^-1

Answer 1

Data : V = 2.5 L

T = 300 K

`pi=0.75atm`

Molar mass of CaCl₂ = M₂ = 111g/mol

R = 0.082 L.atm.k^-1mol^-1.

Solution :

`i=(pi_(obs))/(pi_(cal)`

`2.47=0.75/pi_(cal)`

`pi_(cal)=0.75/2.47`

`pi_(cal)=0.303atm`

`piV=nRT`

`:.piV=W_2/M_2xxRxxT`

`:.pi_(cal)xxV=W_2/M_2xxRxxT`

`:.W_2=(pi_(cal) xxVxxM_2)/(RxxT)`

`=(0.303xx2.5xx111)/(0.082xx300)`

W2=3.417gm

Mass of CaCl₂=3.417gm