1.

Density of 2.05 M solution of acetic acid in water is 1.02g//mL. The molality of same solutionis:

Answer»

1.14 m
3.28 m
2.28 m
0.44 m

Solution :2.05 M solution of acetic acid means that `(2.05xx60=123 G)` of acid is present in 1000 mL of solutiion
Mass of solution=`Vxxd`
=`(1000mL) XX (1.02 g mol^(-1))`
=1020 g
`"Mass os SOLVENT"= 1020-123=897 g.`
`"Molality (m)"=((2.05 mol))/((0.897 KG))`
`=2.28 mol kg^(-1)=2.28 m`


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