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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)`) |
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Answer» 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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