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Find the molality of a solution containing a non-volatile solute if the vapour pressure is 2% below the vapour pressure or pure water. |
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Answer» Solution :`"Accoriding to Raoult's Law "(P_(A)^(@)-P_(S))/P_(S)=n_(B)/n_(A)` `" If" n_(B)" is the number of moles of the solute present in 1000 G of the solvent, then "n_(B)" will represent molality of the solution. LET "P_(A)^(@)=1 ATM, P_(S) =0.98 atm, P_(S)^(@)-P_(S)=0.02 atm, W_(A)=1000g, M_(A)-18G mol^(-1)` `(p_(A)^(@)-P_(S))/P_(S)=(n_(B)xxM_(A))/W_(A)` `n_(B)=((p_(A)^(@)-P_(S)))/P_(S)xxW_(A)/M_(A)=((0.02 atm))/((0.98 atm))xx((1000g))/((18g mol^(-1)))=1.134 mol` The molality (m) of solution is 1.134 m because of the mass of solvent is one kg. |
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