Question

A solution of phosphoric acid was made by dissolving 10.0 g of ${\mathbf{\text{H}}}_{\mathbf{3}}{\mathbf{PO}}_{\mathbf{4}}$ in 100.00 mL of water. The resulting volume was 104 mL. Calculate the density, mole fraction, molarity, and molality of the solution. Assume water has a density of 1.00 g/cm3.

Short answer

Density of the solution$=1.057{\text{g/cm}}^3$

Mole fraction of the solution= 0.0180

Molarity of the solution = 0.980M

Morality of the solution =1.02 m

Step-by-step solution

Step 1:

To find the density of the solution.

We know that density of a solution$=\frac{\mathrm{mass}}{\mathrm{volume}}$

Volume of the solution is given as 104 ml

To find density all we need is to find the mass of the solution which is equal to mass of water plus the mass of phosphoric acid.

Mass of phosphoric acid =10 g ( Given )

Mass of water = density volume = 100 g

Mass of solution=( 100+10 ) g =110 g

Density of solution$=\frac{110}{104}=1.057{\text{g/cm}}^3$

Step 2:

Mole fraction of the component$=\frac{n_a}{n_a+n_b}$

Moles of water$=\frac{\text{given mass}}{\text{molar mass}}$

$$\begin{gathered} =\frac{100}{18} \\ =5.5\text{mol} \end{gathered}$$

Moles of phosphoric acid. $=\frac{10}{97.99}=0.102\text{mol}$

Mole fraction of phosphoric acid$=\frac{0.102}{0.102+5.55}=0.0180$

Step 3:

To find the molarity of the solution

$$\begin{gathered} \text{Molarity}=\frac{\text{moles of solute}}{\text{volume of solution in litres}} \\ =\frac{0.102}{0.104} \\ =0.980\text{M} \end{gathered}$$

:

To find the molality of the solution

$\text{Molarity}=\frac{\text{moles of solute}}{\text{mass of solvent in kilograms}}$

Moles of phosphoric acid = 0.102mol

Weight of solvent in kg = 0.1 kg

$$\begin{gathered} \text{Molarity}=\frac{0.102}{0.1} \\ =1.02 \end{gathered}$$