Question

The density at 20 ∘C of a 0.828 M solution of acetic acid in water is 1.0052 g/mL. The molar mass of acetic acid, CH3CO2H, is 60.05 g/mol. What is the molality of the solution?

Answer 1

Answer : The molality of solution is, 0.866 mole/kg

Explanation :

The relation between the molarity, molality and the density of the solution is,

`d=M[(1)/(m)+(M_b)/(1000)]`

where,

d = density of solution = 1.0052 g/ml

m = molality of solution = ?

M = molarity of solution = 0.828 M = 0.828 mol/L

`M_b` = molar mass of solute (acetic acid) = 60.05 g/mole

Now put all the given values in the above formula, we get the molality of the solution.

`1.0052g/ml=0.828mol/L* [(1)/(m)+(60.05g/mole)/(1000)]`

`m=0.866mol/kg`

Therefore, the molality of solution is, 0.866 mole/kg