Chemistry, 28.08.2019 16:00 smkw04p3ao0n

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

Answers: 1

Show answers

Answers

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).

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

Therefore the molality of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ solution is 3.47 m.

Learn more:

1. What is the concentration of alcohol in terms of molarity?

2. What is the molarity of the stock solution?:

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Solutions

Keywords: concentration terms, molality, m, mol/kg, H3PO4, moles of H3PO4, mass of H2O, H2O, 25.4 %, 25.4 g, 100 g, 74.6 g, 0.2592 mol, 3.47 m.

Answer from: Quest

answer; /// the amount of material dissolved in a measure of liquid(concentration) the above statement is (true); /// good luck

Answer from: Quest

6.4 cm (max) - 25 cm (min)

Another question on Chemistry

Chemistry, 22.06.2019 00:30

What must happen before a body cell can begin mitotic cell division

Answers: 2

Answer

Chemistry, 22.06.2019 03:50

Which of the following statements about acidic water is true? a. acid has no effect on the h,o molecules. b. the solution contains a larger number of oh ions than h,o ions. c. the solution contains a larger number of h,o ions than qh ions. d. the solution contains an equal number of h,o ions and oh ions. none of the above e.

Answers: 1

Answer

Chemistry, 22.06.2019 22:00

The volume of an unknown substance in a sealed glass jar is 50 milliliters. the volume of the jar is 200 milliliters. which state of matter could the substance be?

Answers: 2

Answer

Chemistry, 23.06.2019 05:40

Which order shows the levels of organization from largest to smallest? organism, organ system, cell, organ, tissue organism, tissue, organ system, organ, cell organism, organ, organ system, cell, tissue organism, organ system, organ, tissue, cell

Answers: 2

Answer

You know the right answer?

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

Questions