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The osmotic presure of a urea solution is 500 mm of Hg at 10^(@)C. The solution is diluted and its temperature is reised to 25^(@)C. It is now found that the osmotic pressure of the solution is reduced to 105.3 mm. Detyermine the extrant of dilution of the solution. |
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Answer» `piV=n_(B)RT or pi=(n_(B)RT)/V` `At 10^(@)C:""pi=((500MM))/((750mm))xx(1"atm")=500/760"atm",T=10+273=233=283 K` `(500/760"atm")=(n_(B)xxRxx283K)/V` `At20^(@)C:""pi=(10.5.3mm)/(760mm)xx1(atm)=105.3/760("atm"),T=20+273=293K` `((105.3)/760"atm")=(n_(B)xxRxx293K)/V'` `"Dividing eqn. (i) by eqn.(II),"(500/760"atm")xx(760/105.3"atm"^(-1))=((283K))/Vxx(V')/((293K)` `(V')/V=500/105.3xx((293K))/((283K))=4.92~~5.` V'=5V This means that the solution has been diluted five times. |
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