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A solution containing 7.45 g of KCI per litre of solution has an osmotic pressure of 4.68 atm 300 K. Calculate the degree of dissociation of KCI in solution. |
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Answer» `"According to Van't Hoff equation ":M_(B)=(W_(B)RT)/(PIV)` `W_(B)=7.45g, pi=4.68 atm, V=1 L, T=300 K, R= 0.0821" L atm K"^(-1)mol^(-1)` `M_(B)=((7.45g)xx(0.0821"L atm K"^(-1)mol^(-1))xx(300K))/((4.68" atm")xx(1L))=39.2" g mol"^(-1)` Step II. Calculation of Van't Hoff factor `i=("NORMAL molar mass")/("Observed molar mass")=((74.5"g mol"^(-1)))/((39.2"g mol"^(-1)))=1.9` Step III. `"Calculation of degree of dissociation "(alpha)` `alpha=(i-1)/(n-1)=(1.9-1)/(2-1)=0.9=0.9xx100=90%`. |
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