Question

Calulate molality of 2.5 g of ethanoic acid in 75g of benzene

Answer 1

Molecular wt of ethanoic acid = 46
 Molality = wt. of solute x 1000 / molecular wt.of solute x wt, of solvent in grams
     
              = 2.5 x 1000 / 46 x 75 = 0.72m

Answer 2

Answer : The molality is, 0.55 mole/Kg

Solution : Given,

Mass of benzene = 75 g

Mass of ethanoic acid = 2.5 g

Molar mass of ethanoic acid = 60 g/mole

Molality : It us defined as the number of moles of solute present in one kilogram of solvent.

Formula used :

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent in Kg}}$

In this question, solute is ethanoic acid and solvent is benzene.

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

$Molality=\frac{2.5g\times 1000}{60g/mole\times 75g}=0.55mole/Kg$

Therefore, the molality is, 0.55 mole/Kg