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The vapour pressure 2.1% of an aqueous ...

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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The correct Answer is:
58.3 g `mol^(-1)`
`(P_(A)^(@)-P_(S))/P_(S)=x_(A)`
`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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