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# Mass Concentration

01st Dec 2019 @ 10 min read

Analytical Chemistry

Mass concentration—like molar concentration, molality, mass fraction —is a quantity to measure the concentration of a solute in the solution. It is the ratio of the mass of a solute present in the solution to the volume of the solution. It tells us how much a solute is present in a given volume of the solution. Let try to understand it will an example of coffee.

A simple instant coffee is prepared by mixing the coffee and sugar in hot water. For a typical cup (250 mL) of coffee, we need 10 g of the coffee and 5 g of sugar. The amount could vary based on personal choice. For preparing two cups (500 mL), we will need twice the mass of the ingredients—20 g of coffee and 10 g of sugar. Similarly, we will need 100 g of the coffee and 50 g of sugar for ten cups.

Ingredients in coffee
Cup of coffee Mass of the coffee Mass of sugar Mass of the coffee per cup Mass of sugar per cup
1 (250 mL)10 g5 g10 g5 g
2 (500 mL)20 g10 g10 g5 g
3 (750 mL)30 g15 g10 g5 g
10 (2500 mL)100 g50 g10 g5 g

From the above table, we can see the mass of each ingredient per cup (250 mL) remains constant. In chemistry, we call it mass concentration, the mass of an ingredient per unit volume.

## Definition

The mass concentration of a solute is defined as the mass of the solute per unit volume of the solution.

## Formula

The formula of mass concentration is as follows. Here, ρi is the mass concentration of the i-th solute, mi is the mass of the i-th solute, and V is the volume of the solution.

Note: The notation for mass concentration as well as density is ρ. This may confuse. Both quantities can be distinguished from each other by using a subscript, for example, ρi for mass concentration and ρ for density.

## Units

The SI unit of mass concentration is kg m−3. The equivalent units of kg m−3 are g L−1, g dm−3, and mg mL−1 Other common units are µg mL−1, µg L−1, and µg m−3. ## Explanation

As per the above definition, the ratio of the mass of a solute to the volume of the solution is mass concentration. In chemical laboratories, chemicals are usually measured in grams and the volume of solution in mL or L. Thus, we can redefine the definition as grams of a solute per mL (or L) of the solution. It is a more practical definition. 0.90 g L−1 of NaCl solution contains 0.90 g of NaCl in 1 L of the solution. In 20 g L−1 solution, there will be 20 g of NaCl in 1 L. Thus, as the concentration of a solute increases its mass per litre also increases. The mass of solute in a given volume increases with the concentration.

#### Example 1: Potassium hydroxide

Statement: Potassium hydroxide (KOH) is a strong base. It is soluble in water and have many industrial uses particularly in soap industries.

121 g of KOH is dissolved in water. The volume of the solution is 100 mL. Find the concentration of KOH?

Solution: The mass of KOH is 121 g. mKOH = 121 g. Let ρKOH be the concentration of KOH in the solution. The concentration of KOH is 1.21 g mL−1.

Note: Solubility is often expressed in mass concentration. In the above example, 1.21 g mL−1 is the solubility of KOH in water at 25 °C.

#### Example 2: Liquefied petroleum gas

Statement: Liquefied petroleum gas (LPG) is a flammable liquid mixture of propane and butane. It is commonly used as fuel and sold as cylinders for domestic needs.

An LPG cylinder consists of 40 % propane and 60 % butane. Find the mass concentration of propane if the density of the mixture is 0.58 g mL−1?

Solution: Assume 100 g of a mixture of propane and butane. It contains 40 g of propane and 60 g of butane. The density of the LPG is 0.58 g mL−1. The volume of the mixture is mass divided by density. Let ρC3H8 be the concentration of propane in the mixture. ## Medical science

Mass concentration is very useful in medicine. The composition of liquids drugs is usually mentioned in mass to volume (m/v) percentage i.e. the mass of drug present in a given volume of solution. These drugs are mainly stored in IV bags (see the image below).

Some examples of such solutions are as follows:

1. Isotonic saline: It is 0.9 % NaCl. It contains 0.9 g of the salt in 100 mL of solution.
2. 5 % Dextrose: It has glucose as the solute. It consists of 5 g of glucose in 100 mL of solution.
3. Dextrose saline: It has glucose as well as salts.

Note: In biology, the symbol % is misused. 1 % means 1 g of a solute dissolved in 100 mL of the solution.

#### Example 3: Saline solution

Statement: In a patient, 0.9 % of NaCl is administered. The volume of the solution injected is 250 mL. Find the mass of NaCl injected in the body?

Solution: In biology, 0.9 % means 0.9 g of NaCl in 100 mL of the solution. Thus, the concentration of NaCl is 0.009 g mL−1.

The volume of the solution injected in the patient's body is 250 mL. Let min be the mass of NaCl injected. ## Properties of mass concentration

1. It is a property of a solution.
2. It is an intensive property since the ratio of two extensive properties always gives an intensive property. As a result, it will not vary from sample to sample.
3. It is a concentration based on volume. Hence, it depends on the volume of solution.
4. Since the volume is affected by pressure and temperature, mass concentration is dependent on pressure and temperature.

## Relation with other quantities

### Temperature

As temperature increases, the volume of a solution also increases. This dependence is connected by the thermal expansion coefficient of the solution. Here, V is the volume at T, V0 is the volume at T0, and α is the thermal expansion coefficient of the solution.

Let ρ(i, T) and ρ(i, T0) be the concentration at temperature T and T0. The relation is as follows: ### Density

Let ρi be the density of a solution. The density of a solution is the sum of the mass of each solute divided by the volume of the solution. Therefore, the sum of the concentration of each solute is the density of the solution.

### Mass fraction

Mass fraction is the ratio of the mass of a solute to the total mass and denote by wi. The relation between both is as follows: ### Mole fraction

Mole fraction is the ratio of the mole of a solute to the total moles. The symbol xi is used to represent it. It is related to mass concentration by the following equation. Here, Mi is the molar mass of the solute.

### Molarity

Molarity, aka molar concentration, is similar to mass concentration. It is the ratio of moles of solute, instead of the mass of a solute, to the volume of the solution. ci is molarity of the solute i. ### Molality

Molality (bi) is moles of a solute per kg of the solvent. For n-solute solution, the molality of i-th solute is given by the equation below. Here, M0 is the molar mass of the solvent.

### Equivalent ratios

Let i and j be any two components in a solution. The relationship between each of the above quantities is as follows: ## Solved problems

### Problem 1: Earth's atmosphere

Statement: Earth's atmosphere is a mixture of various gases. It extends up to an altitude of 90 km. Oxygen and nitrogen are dominant gases in the atmosphere. The mole percentage of oxygen and nitrogen is 21 % and 79 %. Find the mass concentration of each gas if the molar mass of oxygen and nitrogen is 32 g mol−1 and 28 g mol−1?

Solution: In 100 mol of air, the amount of oxygen and nitrogen is 21 mol and 79 mol respectively.

Calculating the mass of oxygen and nitrogen in 100 mol of air,  According to the ideal gas law, 1 mol of air occupies 22.4 dm−3 at STP (P = 101.325 kPa and T = 273.15 K). For 100 mol of air, the volume will be 2240 dm−3. Thus, V = 2240 dm−3 = 2240 L.

The mass concentration of oxygen and nitrogen are calculated as follows:  If you add both concentrations, you get the density of air at STP. ### Problem 2: Glucose

Statement: Glucose (C6H12O6) is a monosaccharide sugar. It is an primary energy source not only for humans but also for most organisms. Glucose is produced by plants through photosynthesis. It the most abundant monosaccharide and found in our blood too.

The solubility of glucose at 25 °C is 900 g L−1. Find molecules of sugar present in 100 mL of saturated glucose solution?

Solution: The concentration of glucose is 900 g L−1. Calculating the mass of glucose in 100 mL (= 0.100 L) of the solution, The molar mass of glucose is 180 g mol−1. Moles of glucose in 90.0 g is as follows: One mole of glucose has molecules equal to Avogadro's number NA. Molecules in 0.500 mol of glucose is 0.500 × NA. ### Problem 3: Lithium chloride

Statement: Lithium chloride (LiCl) is a white ionic compound with high solubility in polar solvents like water. It is primarily used for production of lithium, which is a key component in electronic batteries.

Find the concentration of Li+ ions and Cl ions when 50 g of LiCl is dissolved in 100 mL of water?

Solution: The molar mass of LiCl is 42.4 g mol−1. Let the number moles of LiCl in 42.4 g mol−1 be nLiCl. One mole of LiCl contains one mole of Li+ and one mole of Cl. Accordingly, 1.18 mol of LiCl will have 1.18 mol of Li+ and 1.18 mol of Cl.

Calculating the mass of Li+ and Cl,  Note: mLi+ + mLi+ = mLiCl since mass is always conserved.

The volume of the solution is approximately equal to the volume of water, which is 100 mL. V = 100 mL.

Calculating the concentration of Li+ and Cl,  ## Practice problems

Problem 1: Find the concentration of NaOH in a solution if 10 moles of NaOH is dissolved in the 1.00 L of water? The molar mass of sodium hydroxide is 40 g mol−1

Problem 2: The molar concentration of sulphuric acid is 4.00 mol L−1. Find the mass concentration of H2SO4 and SO
4
? The molar mass of H2SO4 and SO
4
is 98.1 g mol−1 and 96.1 g mol−1.

Problem 3: Aluminium hydroxide (Al(OH)3) is used as an antacid—a chemical used to treat acidity. A patient takes two teaspoons (5 mL each) of syrup. The concentration of aluminium hydroxide in the syrup is 50 mg mL−1. Find the intake of aluminium hydroxide?

1. The concentration of NaOH is 400 g L−1.
2. The mass concentration of H2SO4 and SO
4
is 392 g L−1 and 384 g L−1.
3. The intake of Al(OH)3 is 500 mg.

#### Summary

• Mass concentration is the mass of a solute per unit volume of the solution.
• It is denoted by ρi.
• Its formula is .
• The SI unit is kg m−3.
• It is an intensive property. But it may vary with pressure and temperature.

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