Question

Calculate the molality of a solution that is made up of 250 g of water and 20.0 g of NaOH.

Answer 1

m(H₂O)=250 g=0.250 kg
m(NaOH)=20.0 g
M(NaOH)=40.0 g/mol

n(NaOH)=m(NaOH)/M(NaOH)

Cm=n(NaOH)/m(H₂O)

Cm=m(NaOH)/{M(NaOH)m(H₂O)}

Cm= 20.0/{40.0*0.250}=2 mol/kg

Answer 2

Answer : The molality of a solution is, 2 mole/Kg

Explanation : Given,

Mass of solute (NaOH) = 20 g

Mass of solvent (water) = 250 g

Molar mass of NaOH = 40 g/mole

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

Formula used :

`\text{Molality}=\frac{\text{Mass of }NaOH* 1000}{\text{Molar mass of }NaOH* \text{Mass of water}}`

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

`\text{Molality}=(20g* 1000)/(40g/mole* 250g)=2mole/Kg`

Therefore, the molality of a solution is, 2 mole/Kg