Question

An aqueous solution that is 50.0 percent sulfuric acid (h2so4) by mass has a density of 1.143 g/ml. determine the molality of the solution.

Answer 1

Molality is defined as the ratio of number of moles of solute to the mass of solvent in kilograms.

The formula of molality is:

`Molality = (number of moles of solute)/(mass of solvent in kg)` - (1)

The formula of density is:

`Density = (mass)/(volume)` - (2)

Consider,
`Volume = 1 L = 1000 mL`

Rearranging the formula (2):

`mass of solution = density * volume`

Substituting the values:

`mass of solution = 1.143 g/mL * 1000 mL = 1143 g`

Since, the aqueous solution is 50% by mass that is:

`50 = (mass of H_2SO_4)/(mass of solution)* 100`

Rearranging the equation:

`mass of H_2SO_4 = (50)/(100)* mass of solution`

`mass of H_2SO_4 = (50)/(100)* 1143 = 571.5 g`

Now, for determining the number of moles of
`H_2SO_4`:

`n_(H_2SO_4 ) = (mass of H_2SO_4 )/(Molar mass of H_2SO_4 )`

`n_(H_2SO_4 ) = (571.5 g)/(98.08 g/mol) = 5.83 moles`

`mass of solvent = mass of solution - mass of solute`

`mass of solvent = 1143 g - 571.5 g = 571.5 g = 0.5715 kg`

Substituting the values in formula (1):

`molality = (5.83 moles)/(0.5715 kg) = 10.20 m`

Hence, the molality of the solution is
`10.20 m`.