`x g` of non-electrolytic compound (molar mass =200) is dissolved in `1.0 L` of `0.05 M NaCl` solution. The osmotic pressure of this solution is found to be `4.92 atm` at `27^(@)C`. Calculate the value of `x`. Assume complete dissociation of `NaCl` and ideal behaviour of this solution.

Explanation

Given that,
`pi = 4.92 atm, T = 27 + 273 = 300 K, V = 1 litre`
`n_(1) = "mole of non-electrolyte" = (x)/(200)`
`n_(2) = "mole" = NaCl = 0.05`
Using `piV = nST` for a solution containing non-electroyte and `NaCl`
`piV = n_(1)ST + n_(2)(1+alpha)ST`
`piV = [n_(1) + n_(2)(1+alpha)]ST`,
where `alpha = 1`, given for `NaCl`
`:. 4.92 xx 1 = [(X)/(200) + 0.05 xx2] xx 0.082 xx 300`
`:. X = 20 g`