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27th Nov 2019 @ 9 min read
Molality like any other concentrations—molarity, mass concentration, mole fraction—is a way to measure the amount of a solute present in the solution. It is the amount of solute present in the solution per mass of the solvent. Before we discuss it further, let us take an example.
Watermelons are one of the best fruits to eat in hot summer days. They are juicy, delicious, and refreshing. A typical juice content in a watermelon is between 9 kg to 10 kg (20 lb to 22 lb). Watermelons are packed with a large number of seeds. An average American watermelon has 600 seeds. Thus, the number of seeds per kg of juice is around 60.
Here, the number of the seeds is analogous with the amount of a solute and the mass of the juice with the mass of the solvent. And the number of the seeds per kg of the juice is analogous with the molality of a solute.
|Seeds per kg juice||Molality|
The molality of a solute in the solution is defined as the amount of the solute (in moles) per mass of the solvent (in kg). It is also known as the molal concentration of a solute.
The symbol b is used to denote molality.
The standard unit of molality is moles per kilogram (mol kg−1). It also an SI unit. mol kg−1 is also called molal. In other words, 1 molal = 1 mol kg−1.
Note: The unit molal is sometimes abbreviated by the symbol m. But this practice is becoming obsolete and no longer recommended.
The molality of a solute tells the amount of the solute present in the solution per mass of the solvent.
Here, the amount of solute is always expressed in moles and the mass of solvent in kilograms.
0.5 molal of NaOH aqueous solution means 0.5 mol of NaOH is dissolved in 1 kg of water. Similarly, a 1.0 molal of NaOH solution has 1.0 mol of NaOH in 1 kg of water. As molality increases, the amount of solute present in the solution also increases.
Statement: Copper sulphate (CuSO4) is a bluish inorganic compound. It is soluble in water. Copper sulphate is used as a chemical reagent in Fehling's solution and Benedict's solution.
In a CuSO4 solution, CuSO4 and water content is 100 g and 900 g respectively. Find the molal concentration of CuSO4 if the molar mass of CuSO4 is 159.6 g mol−1?
Solution: Let MCuSO4 be the molar mass of CuSO4. MCuSO4 = 159.6 g mol−1.
The moles of CuSO4 in the solution be nCuSO4.
The mass of the solvent, which is water, is 900 g. mH2O = 0.900 kg.Therefore, the molality of CuCO4 is 0.708 molal.
For a solution with more than one solute, let bi be the molality of the i-th solute having ni moles, and m0 is the mass of the solvent.
Let take an example of a solution with more than one solute.
Statement: Seawater is one of the best examples of a solution with multiple solute. There are numerous salts in this giant water body. The presence of these salts makes seawater saline. The most common ions in seawater are chloride (Cl−), sodium (Na+), magnesium (Mg2+), and sulphate (SO2−
A sample of 1 L of seawater contains is 19.4 g of Cl−, 10.8 g of Na+, 1.3 g of Mg2+, and 1.8 g of SO2−
4. Find the molality of each ion if the density of the water is 1.03 kg L−1.
Solution: The molar mass of each of the ions is shown in the table below.
|Ion||Molar mass (g mol−1)|
Calculating moles of each ion,
Now, the mass of the water is volume times density.
The molal concentration of each ion is calculated as follows:
When a solution consists of multiple solvents, the molality of a solute is calculated based on the sum of the mass of each solvent. We can treat this mixture of solvents as a pseudo-solvent.
Here, m0 is the mass of the pseudo-solvent.
Statement: Aqua regia is a mixture of nitric acid (HNO3) and hydrochloric acid (HCl). The molar ratio of HNO3 to HCl is 1:3. It is yellowish fuming liquid, which can dissolve gold.
In the above reaction, 1 mol of Au requires 3 mol of HNO3.
Find the molal concentration of Au when 1.0 mol of Au is dissolved in 3.0 mol of HNO3 and 9.0 mol of HCl?
Solution: The number of moles of solute, which is gold, is 1.0 mol. nAu = 1.0 mol.
The molar of HNO3 and HCl is 63.0 g mol−1 and 36.5 g mol−1.
The total mass of the solvents is mHNO3 + mHCl.
The molality of the solute be bAu.
In all the above examples, we have calculated the molal concentration of a solute in a solution, never of the solvent. Let b0 be the molality of the solvent in a solution and moles of the solvent be n0.
Here, M0 is the molar mass of the solvent.
Consider a binary mixture (one solvent and one solute). All the quantities of the solvent is represented subscript 1 (e.g. m1, b1, ρ1) and for the quantities of the solute with subscript 2 (e.g. m2, b2, ρ2).
The molality of the solute is .
Mass concentration is the ratio of the mass of the solute to the volume of the solution.
Here, V is the volume of the solution. Both quantities are related by the formula below.
Mass fraction is the ratio of the mass of solute to the total mass.
The relation between mass fraction and molal concentration is as follows:
Mole fraction is the ratio of moles of the solute to the total mole.
The molal concentration is expressed in terms of mole fraction by the formula below.
Molarity and molality are often confused with each other. But they are completely different quantities. The former is a volumetric measure while the latter is a mass measure. Molarity is the ratio of moles of the solute to the volume of the solution.
The relation between both is as follows:
Let i and j be any two constituents (solute or solvent) in a solution. The relationship between each of the above quantities is as follows:
Statement: Lemonade is a common beverage found throughout the world. It is a homogeneous mixture of lemon water (≈ 5 % citric acid), water, and sugar.
A glass of lemonade consists of 200 mL of water, 1 g of citric acid, and 20 g of sugar. The molar mass of each are 18.0 g mol−1 192.1 g mol−1, and 342.3 g mol−1 respectively. Find the molality of citric acid and sugar if the density of the water is 1.00 g mL−1?
Solution: Let Msugar and Mcitric be the molar mass of sugar and water. Mcitric = 192.1 g mol−1 and Msugar = 342.3 g mol−1.
Calculating moles of each solute,
Mass is density times volume. Let mwater be the mass of the solvent.
Molality of the solutes is calculated as follows:
Statement: hydrochloric acid (HCl) is one the strongest acid. It is colourless and miscible in water.
Find the molal concentration of a 3 % HCl aqueous solution?
Solution: A 3 % HCl solution consists of 3 % of HCl and 97 % of water. Assume a 100 g of HCl solution. Thus, the mass of HCl is 3 g and water, 97 g.
The molar mass of HCl is 36.5 g mol−1. Calculating moles of HCl,
The mass of solvent (water) is 97 g = 0.097 kg. Therefore, the molal concentration of HCl is as follows:
Statement: Potassium dichromate (K2Cr2O7) is a well-known oxidizing agent. It is dark orange in colour and soluble in water.
A 3.0 molal of K2Cr2O7 solution is required in an experiment. Find the mass of K2Cr2O7 to be dissolved in 200 mL of water? Assume the density of water to be 1.0 g mL−1.
Solution: The volume of the solvent is 200 mL. The mass of the solvent is density times volume.
The molal concentration of K2Cr2O7 is 3.0 molal.
The molar mass of K2Cr2O7 is 294 g mol−1.
The mass of K2Cr2O7 to be dissolved in 200 mL of solution is 176 g.
Problem 1: A NaCl solution is made by mixing 100 g of the salt in 1.0 L of water. Find the molal concentration of NaCl if the density of water is 1.00 g mL−1? The molar mass of NaCl is 58.5 g mol−1.
Problem 2: A mixture contains 10 g of ethanol and 90 g of isopropyl alcohol (solvent). Find the molal concentration of ethanol? The molar mass of ethanol and isopropyl alcohol is 46 g mol−1 and 60 g mol−1.
Problem 3: Find moles of sulphuric acid (H2SO4) in 1.2 molal of H2SO4? The mass of the H2SO4 solution is 150 g.
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