Determine the amount of `CaCl_2` (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at `27^@C` .

Explanation

For a solute undergoing dissociation, the following equation may be used
`pi = iCRT = i n/V RT ` (i is van.t Hoff.s factor)
`pi = (pi xx V)/(i xx R xx T)`
Substituting the values, we get
`n = (0.75 atm xx 2.5L)/(2.47 xx 0.0821 L atm K^(-1) "mol"^(-1) xx 300 K) = 0.0308` mole
Molar mass of `CaCl_2 = 40 + 2 xx 35.5 = 111 g "mol"^(-1)`
` therefore ` Amount dissolved = 0.0308 x 111 g = 3.42 g.