1.

The density of a 2.03 M solution os acetic acid (molecular mass=60) in water is 1.790 g/mL. Calculate the molality of the solution.

Answer» Correct Answer - 2.267 M
`"Strenght of the solution"="Molarity"xx"Molar mass"`
`=(2.03"mol L"^(-1))xx(60 " g mol"^(-1))=121.8 gl^(-1)`
`"Density of the solution"=1.017" g mL"^(-1)`
`"Mass of one litre of the solution"=(1000 mL)xx(1.017" g mL"^(-1))=0.17g`
`"Mass of water in the solution"=1017-121.8=895.2g`
`"Molality (m)"=("No. of moles of acetic acetic acid")/("Mass of solvent in kg")`
`=((2.03mol))/((895.2//1000kg))=2.267" mol kg"^(-1)=2.267 m.`


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