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The vapour pressure 2.1%of an aqueous solution of a non volatile solute at 373 K is 755 mm Hg. Calculate the molar mass of the solute. |
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Answer» `P_(A)^(@)=760" MM hg, "P_(S)=755" mm Hg "` `x_(B)=(P_(A)^(@)-P_(S))/P_(S)=(760" mm Hg "-755" mm Hg ")/(755mmhg)=0.0066` A 2.1% awueous solution means 2.1 g of SOLUTE in 100 g of solution `therefore "Mass of SOLVENT "=100 g-2.1g=97.9g ` `therefore "Mass of solvent" =100G 2.1 g-2.1 g =97.9 g` `"Moles of solvent "n_(A)=(97.9g)/(18" g mol"^(-1))=5.439 g` `x_(B)=n_(B)/(n_(A)+n_(B))" or "1/x_(B)=(n_(A)+n_(B))/n_(B)=n_(A)/n_(B)+1` `n_(A)/n_(B)+1/x_(B)-1=1/(0.0066)-1=151.5-1=150.5` `n_(B)=n_(A)/(150.5)=((5.349mol))/(150.5)=0.036 mol.` `"MOLAR mass of folute"=("Mass of solute")/("Moles of solute")=((2.1g))/(("0.036mol"))=58.3" g mol"^(-1)` |
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