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Aqueous solution of 3% non-volatile solute exerts a pressure of 1 bar at the boiling point of the solvent. Calculate the molar mass of solute. |
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Answer» Solution :DATA: (i) Vapour pressure of water at its [BOILING point temperature]`=P^(@)=1.013` bar (ii) Vapour pressure of solution at normal boiling point P=1bar 3% solution=3g of solute in 100G of solution. (iii) Mass of solute=3g(iv) Mass of solvent=100-3=97g (v) RELATIVE lowering of vapoure pressure`=(P^(@)-P)/(P)=(1.013-1)/(1.013)=0.0128` (vi) Molar mass of water=18g `mol^(-1)` Formula, `M_(2)=(W_(2))/(W_(1))xx(M_(1))/([(P^(@)-P)/(P^(@))])` Substitution: `M_(2)=(3)/(97)xx(18)/(0.0128)` ANSWER `M_(2)=43.49g//mol`. |
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