At `10^@C`, the osmotic pressure of 1% (w/v) solution of 'X' is `7.87 xx 10^(4) Nm^(-2)`. What is the molecular weight of solute X ?

Explanation

Osmotic pressure of the given solution = `pi = 7.87 xx 10^(4) N m^(-2)`
Volume of the solution = `V = 100 mL = 10^(-4) m^(3)`
Solution constant =`S = 8.314 J K^(-1) mol^(-1)`
Temperature = `2.73 + 10 = 283 K` , Weight of solute = w =1g
Molecular weight of solute = `(wST)/(pi V) = (1 xx 8.314 xx 283)/(7.87 xx 10^(-4) xx 10^(-4)) = 299 g mol^(-1). `