Question

What is the molality of phosphoric acid, ${\mathbf{H}}_{\mathbf{3}}\mathbf{P}{\mathbf{O}}_{\mathbf{4}}$, in a solution of 14.5 g of ${\mathbf{H}}_{\mathbf{3}}\mathbf{P}{\mathbf{O}}_{\mathbf{4}}$in 125 g of water?

1. Outline the steps necessary to answer the question.
2. Answer the question.

Short answer

The molality of Phosphoric Acid, ${\mathbf{H}}_{\mathbf{3}}\mathbf{P}{\mathbf{O}}_{\mathbf{4}}\mathbf{=}\mathbf{1}.18\mathbf{m}$

Step-by-step solution

Molality

Molality is the number of moles of solute present in $1.0kg$ solvent.

Subpart (a)

The molality of the solution can be calculated by using the following steps:

i) The number of moles of solute$\left(n_{solute}\right)$

ii) Substituting the values in molality formula:

$Molality\left(m\right)=\frac{n_{solute}}{Mass{}_{solvent}}$

Subpart (b)

The mass of solute $\left(H_{3}P{O}_4\right)$ is 14.5 g.

The molar mass of solute is 98 g/mol.

The number of moles of $\left(H_{3}P{O}_4\right)$is:

$\begin{array}{rl}{n}_{H_{3}P{O}_4}& =\frac{14.5g}{98g/mol}\\ & =0.148mol\end{array}$

The mass of solvent is 125 g that is 0.125 kg.

Substitute these values in molality formula:

$\begin{array}{rl}m& =\frac{n_{solute}}{Mas{s}_{solvent}}\\ & =\frac{0.148mol}{0.125kg}\\ & =1.18mol/kg\end{array}$