Question

Calculate the molality of \(1 \mathrm{~L}\) solution of \(80 \% \mathrm{H}_{2} \mathrm{SO}_{4}(w / V)\), given that the density of the solution is \(1.80 \mathrm{~g} \mathrm{~mL}^{-1}\). (a) \(8.16\) (b) \(8.6\) (c) \(1.02\) (d) \(10.8\)

Short answer

The molality of the solution is approximately 8.16 m.

Step-by-step solution

Calculate the mass of the solution

Calculate the mass of the solution using its volume and density. The mass of the solution (in grams) can be found by multiplying its volume (in mL, noting that 1 L = 1000 mL) by the density in g/mL: Mass of solution = Volume of solution × Density = 1000 mL × 1.80 g/mL = 1800 g.

Calculate the mass of solute (H2SO4)

Calculate the mass of the solute using the percentage concentration (w/v). Since the solution is 80% H2SO4 by weight/volume, this means 80 grams of H2SO4 are present in every 100 mL of solution. So in 1000 mL, there will be: Mass of solute = (80 g/100 mL) × 1000 mL = 800 g.

Calculate the mass of the solvent (water)

The mass of the solvent can be found by subtracting the mass of the solute from the mass of the solution: Mass of solvent = Mass of solution - Mass of solute = 1800 g - 800 g = 1000 g.

Convert the mass of solvent to kilograms

Convert the mass of the solvent from grams to kilograms to use in calculating molality: Mass of solvent in kg = 1000 g ÷ 1000 g/kg = 1 kg.

Calculate the number of moles of solute (H2SO4)

Calculate the number of moles of H2SO4 using its molar mass. The molar mass of H2SO4 is 2(1) + 32 + 4(16) = 98 g/mol. Thus, the number of moles of H2SO4 is: Number of moles = Mass of solute / Molar mass = 800 g / 98 g/mol ≈ 8.1633 moles.

Calculate molality of the solution

Molality is defined as the number of moles of solute per kilogram of solvent. Using the mass of the solvent in kg and the moles of the solute, calculate the molality: Molality = Number of moles of solute / Mass of solvent in kg = 8.1633 moles / 1 kg ≈ 8.16 m.

Key concepts

Chemistry Concentration Units

Understanding concentration units is crucial for chemists to precisely communicate the composition of a solution. One of these units, molality, measures the number of moles of a solute per kilogram of the solvent.

In contrast to molarity, which depends on the volume of the solution, molality is temperature-independent since it relies on mass. This makes it a valuable unit for reactions sensitive to temperature changes. In the workbook exercise, we converted the percentage concentration (80% H2SO4 w/v) to molality by following a step-by-step procedure, involving the calculation of the mass of solute, the mass of the solvent, and finally, the number of moles of solute.

Solution Preparation

The process of creating a solution with a desired concentration involves several steps that need to be performed with care and precision. The starting point is to understand the type of concentration unit provided, such as weight/volume percentage in our exercise.

Preparing a solution then requires measuring the correct amount of solute and solvent, which in our case, was determined by converting the volume of the solvent to its mass using the given density. When doing this in a laboratory, a balance would be used for solid solutes, while graduated cylinders or volumetric flasks would measure solvents. By understanding the components and their interactions, students can better grasp the concept of solution preparation.

Molar Mass

The molar mass is a fundamental concept in chemistry, noted as the mass of one mole of a particular substance. It is typically expressed in grams per mole (g/mol). Molar mass links the macroscopic world of grams with the microscopic world of moles, which represent the number of particles.

In our exercise, the molar mass of H2SO4 was calculated using the atomic masses of hydrogen (H), sulfur (S), and oxygen (O) taken from the periodic table. The importance of understanding molar mass cannot be overstated as it is essential for converting between mass and moles of a substance. This conversion is the bridge to finding the molality of a solution, as we have demonstrated in the example provided.