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

We know that,
`pi = i(n)/(V) RT`
`rArr pi =i(w)/(MV) RT rArr w = (piMV)/(iRT)`
`pi = 0.75 atm`
`V = 2.5 L`
`i= 2.47`
`T =(27 +273)K = 300 K`
Here,
`R =0.082L atm K^(-1)mol^(-1)`
`M = 1 xx 40 +2 xx 35.5 = 111g mol^(-1)`
Therefore, `w = (0.75 xx 111 xx 2.5)/(2.47 xx 0.0821 xx 300) = 3.42g`
Hence, the required amount of `CaCI_(2)` is `3.42g`.