Determine the amount of `CaCl_(2)(i=2.47)` dissolved in 2.5 litre of waster such that its osmotic pressure is 0.75 atm `27^(@)C` at.

Explanation

`pi = (n)/(V)RT`
`pi=i(w)/(MV)RT`
`k=(pi MV)/(iRT)`
`pi-0.75` atm
V = 2.5 L
`i=2.47`
`T=(27+273)K`
= 300 K
Here,
`R=0.0821 " atm K"^(-1)mol^(-1)`
`M=1xx40+2xx35.5`
`= 111 g mol^(-1)`
Therefore,
`w=(0.75xx111xx2.5)/(2.47xx0.0821xx300)`
= 3.42 gram.