Question

The density of 2.05M acetic acid in water is 1.02 g/ cm^3.calcupllate molality of solution
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Answer 1

Answer: This may help you.....

Explanation:

Suppose the volume of solution is 1000 ml

1 ml=1.2 gm

1000 ml=1020 gm

Number ofr moles =2⋅05M×1 lit=2.05 mol

Mass of solute =n×M⋅wt=2.05×60=123 gm

Solvent mass =1020−123=897 gm

Molality =  

60

123

​  

×  

897

1000

​  

=2.28 mol kg  

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Answer 2

Answer:

Molality = 17 molal solution

Explanation:

The density of acetic acid = 1.02 g/cm³

Molarity of acetic acid  = 2.05 M

let volume of solution be 1000cm³

1 cm³ = 1.02 g

1000cm³ = 1.02 x 1000

               = 1.02 x 10³ g

weight of 1000cm³ = 1kg                                   { 1000cm³ = 1L}

no of moles =$\frac{Given Mass}{Molar Mass}$

                   = 1.02 x 10³ g = 17 moles

                       60 g/mol

Molality = no of moles / weight of solvent (kg)

             = 17 mol / 1 kg

Molality = 17 molal solution

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