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The osmotic pressure of a solution containing `5 g` of substance (molar mass =100) in 308 m L of solution was found to be 4.0 atm at `300 K`. Calculate the value of solution constant (R) |
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Answer» The given values are `W_(B)=5 g`,`Mw_(B)=100 mL^(-1)` `pi=4.0 atm`,`T=300 K`,`V=380 mL` Using formula, `R=piV/piT` `=4.0xx(308/1000)/(5/100)xx300=0.0821 L-"atm" "mol"^(-1) "degree"^(-1)` |
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