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`
`=7.87xx10^4 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)/(piV)`
`=(1xx8.314xx283)/(7.87xx10^4xx10^(-4))=299 "g mol"^(-1)`