Question

A solution of phosphoric acid was made by dissolving 10.0 g of H3PO4 in 100.0 mL of water. The resulting volume was 113 mL. Calculate the density, mole fraction, molarity, and molality of the solution. Assume water has a density of 1.00 g/cm3.

Answer 1

Step-by-step explanation:

Mass of solute = 10.0 g

mass of solvent(water) = m

Volume of solvent( water) = v = 100.0 mL

Density of water= d =
`1 g/cm^3=1 g/mL`

`1 mL= 1 cm^3`

`m=d* v=1.0 g/mL* 100.0 = 100.0 g`

Mass of solution(M) = Mass of solute + mass of solvent

M = 10.0 g + 100.0 g = 110.0 g

Volume of the solution = V = 113 mL

Density of the solution = D

`D=(M)/(V)=(110.0 g)/(113 mL)=0.9734 g/mL`

The density of the solution is 0.9734 g/ml.

Moles of phosphoric acid =
`n_1=(10.0 g)/(98 g/mol)=0.1020 mol`

Moles of water =
`n_2=(100.0 g)/(18g/mol)=5.556 mol`

Mole fraction of phosphoric acid =
`\chi_1`

`\chi_1=(n_1)/(n_1+n_2)=(0.1020 mol)/(0.1020 mol+5.556 mol)`

`\chi_1=0.01803`

Mole fraction of water =
`\chi_2`

`\chi_2=(n_2)/(n_1+n_2)=(5.556 mol)/(0.1020 mol+5.556 mol)`

`\chi_2=0.9820`

`[Molarity]=\frac{\text{Moles of solute}}{\text{Volume of solution(L)}}`

Moles of phosphoric acid = 0.1020 mol

Volume of the solution = V = 113 mL = 0.113 L ( 1 mL = 0.001 L)

Molarity of the solution :

`=(0.1020 mol)/(0.113 L)=0.903 M`

`[Molality]=\frac{\text{Moles of solute}}{\text{Mass of solvent(kg)}}`

Moles of phosphoric acid = 0.1020 mol

Mass of solvent(water) = m =100.0 g = 0.100 kg ( 1 g = 0.001 kg)

Molality of the solution :

`=(0.1020 mol)/(0.100 kg)=1.02 mol/kg`