Chemistry, 28.08.2019 16:00 smkw04p3ao0n

Calculate the molality of a 25.4% (by mass) aqueous solution of phosphoric acid (h3po4).

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The molality of a 25.4 % (by mass) aqueous solution of phosphoric acid $\left({{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}}\right)$ is $\boxed{{\text{3}}{\text{.47 m}}}$ .

Further Explanation:

The proportion of substance in the mixture is called concentration. The most commonly used concentration terms are as follows:

1. Molarity (M)

2. Molality (m)

3. Mole fraction (X)

4. Parts per million (ppm)

5. Mass percent ((w/w) %)

6. Volume percent ((v/v) %)

Molality is equal to the amount of solute that is dissolved in one kilogram of the solvent. It is denoted by m and its unit is moles per kilograms (mol/kg). The component present in smaller quantity is solute while that in larger quantity is known as a solvent. The solute gets itself dissolved in the solvent.

The formula to calculate the molality of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ solution is as follows:

${\text{Molality of }}{{\text{H}}_3}{\text{P}}{{\text{O}}_4}\;{\text{solution}}=\frac{{{\text{Amount}}\;\left({{\text{mol}}} \right)\;{\text{of}}\;{{\text{H}}_3}{\text{P}}{{\text{O}}_4}}}{{{\text{Mass}}\;\left({{\text{kg}}}\right)\;{\text{of}}\;{{\text{H}}_2}{\text{O}}}}$ …… (1)

Consider 100 g to be the mass of the sample. Phosphoric acid $\left({{{\text{H}}_3}{\text{P}}{{\text{O}}_4}}\right)$ is 25.4 % by mass. So the mass of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ is 25.4 g.

The formula to calculate the mass of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ solution is as follows:

${\text{Mass of }}{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}{\text{ solution}}={\text{Mass of }}{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}+{\text{Mass of }}{{\text{H}}_{\text{2}}}{\text{O}}$ …… (2)

Rearrange equation (2) to calculate the mass of ${{\text{H}}_2}{\text{O}}$ .

${\text{Mass of }}{{\text{H}}_{\text{2}}}{\text{O}}={\text{Mass of }}{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}{\text{ solution}}-{\text{Mass of }}{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ …… (3)

The mass of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ solution is 100 g.

The mass of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ is 25.4 g.

Substitute these values in equation (3).

$\begin{aligned}{\text{Mass of }}{{\text{H}}_{\text{2}}}{\text{O}}&={\text{100 g}}-{\text{25}}{\text{.4 g}}\\&={\text{74}}{\text{.6 g}}\\\end{aligned}$

Mass of ${{\text{H}}_2}{\text{O}}$ is to be converted into kg. The conversion factor for this is,

${\text{1 g}}={10^{ - 3}}\;{\text{kg}}$

So the mass of ${{\text{H}}_2}{\text{O}}$ is calculated as follows:

$\begin{aligned}{\text{Mass of }}{{\text{H}}_{\text{2}}}{\text{O}}&=\left({{\text{74}}{\text{.6 g}}}\right)\left({\frac{{{{10}^{ - 3}}\;{\text{kg}}}}{{{\text{1 g}}}}}\right)\\&=0.074{\text{6 kg}}\\\end{aligned}$

The formula to calculate the moles of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ is as follows:

${\text{Moles of }}{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}=\frac{{{\text{Given mass of }}{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}}}{{{\text{Molar mass of }}{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}}}$ …… (4)

Substitute 25.4 g for the given mass and 97.99 g/mol for the molar mass of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ in equation (4).

$\begin{aligned}{\text{Moles of }}{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}&=\left( {{\text{25}}{\text{.4 g}}}\right)\left({\frac{{{\text{1 mol}}}}{{{\text{97}}{\text{.99 g}}}}}\right)\\&=0.259{\text{2 mol}}\\\end{aligned}$

Substitute 0.2592 mol for the amount of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ and 0.0746 kg for the mass of ${{\text{H}}_2}{\text{O}}$ in equation (1).

$\begin{aligned}{\text{Molality of }}{{\text{H}}_3}{\text{P}}{{\text{O}}_4}\;{\text{solution}}&=\frac{{{\text{0}}{\text{.2592 mol}}}}{{{\text{0}}{\text{.0746 kg}}}}\\&=3.4746{\text{6 m}}\\&\approx 3.4{\text{7 m}}\\\end{aligned}$