1.

Calculate the molality and mole fraction of 2.5 g of ethanoic acid (CH_(3)COOH) in 75 g of benzene.

Answer»

Solution :Mass of solute `(CH_(3)COOH)="2.5 G,Mass of solvent "(C_(6)H_(6))="75 g = 0.075 KG"`
`"Molar massof "CH_(3)COOH="60 g mol"^(-1),"Molar mass of "C_(6)H_(6)="78 g mol"^(-1)`
Calculation of molality : Moles of the solute `(CH_(3)COOH)=("2.5 g")/("60 g mol"^(-1))=0.0417`
`"Molality"=("Moles of the solute")/("Mass of the solvent in kg")=("0.0417 mol")/("0.075 kg"="0.556 mol kg"^(-1)`
Calculation of MOLE fraction : Moles of solute `(n_(CH_(3)COOH))="0.0417 calculated above"`
`"Moles of solvent "(n_(C_(6)H_(6)))=("75 g")/("78 g mol"^(-1))=0.961`
Mole fraction of `CH_(3)COOH` in the solution `=(n_(CH_(3)COOH))/(n_(CH_(3)COOH)+n_(C_(6)H_(6)))=0.0416`


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