Question

A 1.37 M aqueous solution of citric acid (${\mathbf{H}}_{\mathbf{3}}{\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{5}}{\mathbf{O}}_{\mathbf{7}}$) has a density of $\mathbf{1}\mathbf{.}\mathbf{10}\mathbf{}\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$. What are the mass percent, molality, and mole fraction of the citric acid?

Short answer

Mass percent of citric acid = 23.9%

Molality of citric acid = 1.64m

Mole fraction of citric acid = 0.0286

Step-by-step solution

Mass percent of citric acid

$\text{mass percent =}\frac{\text{grams of solute}}{\text{grams of solution}}\times 100$

We have to assume that the volume of solution is 1L.

1L of the solution has mass = 1000ml density of solution = 1100g

The mass of citric acid = 1.37 molecular mass of citric acid

=1.37192=263g

Therefore mass percent = 23.9%

Molality of citric acid

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

Moles of solute =1.37

Weight of solvent = =0.837kg

molality= 1.64m

Step 3:Mole fraction of citric acid

${\text{mole fraction of componentx}}_a\text{=}\frac{n_a}{n_a+n_b}$

Where$n_a$ =number of moles of component

$n_b$ =number of moles of mixture

Mole of water in the solution= $\frac{837}{18}$=46.5

Therefore nb = 46.5+1.37 = 47.87

Mole fraction of citric acid =$\frac{1.37}{1.37+46.5}$=0.0286