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A solution was prepared by dissolving 6....

A solution was prepared by dissolving `6.0 g` an organic compound in `100 g` of water. Calculate the osmotic pressure of this solution at `298 K`,when the boiling point of the Solution is `100.2^(@)C`. (`K_(b)` for water =`0.52 K m^(-1)`), R=0.082 L atm `K^(_1) mol^(-1)`)

Text Solution

Verified by Experts

The given values are
`W_("solute")=6.0 g`,`W_("solvent")=100 g`,`K_(b)=0.52 K m^(-1)`
`DeltaT_(b)=100.2-100=0.2`
Now using the formula
`Mw_("solute")=(K_(b)xxW_("solute")xx1000)/(DeltaT_(b)xxW_("solvent"))`
`=(0.52xx6.0xx1000)/(0.2xx100)`
`=156 g mol^(-1)`
Now, `pi=(nRT)/(V)`,`n=W_("solute")/(Mw_("solute"))`
`:.n=6.0/156`
`:. n=(6.0xx0.082xx298)/(156xx0.1)`
`=9.398 atm`
Thus, the osmotic pressure of the solution is `9.398 atm`.
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