Question

the density of 2.03 mole solution of Acetic acid is 1.017 gram per mole calculate its molality!?

Answer 1

Given solution contains 2.03 moles of acetic acid.
since molecular mass of acetic acid is 60g
then 2.03 moles = 2.03 × 60g = 121.80 g

In the given 2.03M solution 121.80 g acetic acid is dissolved in 1 L of solution. So we have to find out the weight of 1 L solution.

density of the solution is given 1.017 g/mL
Density = weight / volume
1.017 = wt / 1ml
So, wt = 1.017

1ml of solution weighs 1.017g
so 1000ml will weigh = 1.017× 1000 = 1017g

Now we know the weight of solution. But to calculate the molality we need to know the weight of solvent.
weight of solvent = weight of solution - weight of solute
=1017- 121.80
=895.2g = 0.8952kg
now we can calculate the molality of solution

molality = 2.03moles
0.8952kg
Answer = 2.26m

Answer 2

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