Question

13.66 What is the molality of a solution consisting of \(44.0 \mathrm{~mL}\) of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6} ; d=0.877 \mathrm{~g} / \mathrm{mL}\right)\) in \(167 \mathrm{~mL}\) of hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right.;\) \(d=0.660 \mathrm{~g} / \mathrm{mL}) ?\)

Short answer

The molality is approximately 4.482 mol/kg.

Step-by-step solution

- Calculate the mass of benzene

First, find the mass of benzene using its volume and density. Use the formula mass = density × volume.Given: Volume of benzene = 44.0 mLDensity of benzene = 0.877 g/mLMass of benzene = 0.877 g/mL × 44.0 mL = 38.588 g

- Calculate the mass of hexane

Next, find the mass of hexane using its volume and density.Given:Volume of hexane = 167 mLDensity of hexane = 0.660 g/mLMass of hexane = 0.660 g/mL × 167 mL = 110.22 g

- Convert the mass of hexane to kilograms

Molality requires the mass of the solvent (hexane) in kilograms.Mass of hexane in kg = 110.22 g × (1 kg / 1000 g) = 0.11022 kg

- Calculate the moles of benzene

Find the moles of benzene by using its molar mass.Given:Molar mass of benzene (C₆H₆) = 6 (12.01 g/mol) + 6 (1.01 g/mol) = 78.12 g/molMoles of benzene = Mass of benzene / Molar mass of benzene = 38.588 g / 78.12 g/mol Moles of benzene ≈ 0.494 mole

- Calculate the molality of the solution

Find the molality of the solution by dividing the moles of solute (benzene) by the kilograms of solvent (hexane).Molality (m) = Moles of benzene / Mass of hexane in kg = 0.494 mole / 0.11022 kg Molality ≈ 4.482 mol/kg

Key concepts

Density and Volume Relationship

Understanding the relationship between density and volume is crucial in chemistry, especially when dealing with liquids. Density is defined as the mass per unit volume of a substance. The formula used to calculate mass based on density and volume is: \( \text{mass} = \text{density} \times \text{volume} \).

To put it simply, if you know how dense a substance is and how much volume you have, you can figure out the mass. For example, in the given problem, we calculated the mass of benzene by multiplying its density (0.877 g/mL) with its volume (44.0 mL), resulting in 38.588 grams.

Similarly, with hexane, the mass was found by using its density (0.660 g/mL) and its volume (167 mL), giving us 110.22 grams. Understanding this relationship helps in various scientific calculations and real-world applications.

Mass to Moles Conversion

Converting mass to moles is an important concept in chemistry for understanding the amount of substance present. The number of moles of a substance is calculated by dividing the mass by the molar mass. The molar mass is the mass of one mole of a substance and is usually expressed in grams per mole (g/mol). The formula is: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \).

In our exercise, the molar mass of benzene (C₆H₆) is calculated from the atomic masses of carbon (12.01 g/mol) and hydrogen (1.01 g/mol). With six atoms of carbon and six atoms of hydrogen, the molar mass of benzene is 78.12 g/mol.

By dividing the mass of benzene (38.588 g) by its molar mass (78.12 g/mol), we get approximately 0.494 moles of benzene. This conversion is essential because many chemical calculations, like determining solution concentration, are based on the number of moles rather than the mass.

Solution Concentration Calculations

Calculations involving the concentration of solutions are fundamental in chemistry. One common measure of solute concentration is molality. Molality (m) is defined as the number of moles of solute per kilogram of solvent. The formula for molality is: \( \text{molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \).

In our problem, we determined the number of moles of benzene (0.494 moles) and the mass of hexane (the solvent) in kilograms (0.11022 kg). By dividing the moles of benzene by the kilograms of hexane, we obtain the molality: \( \text{molality} \frac{0.494 \text{ moles}}{0.11022 \text{ kg}} \text{ ≈ 4.482 mol/kg} \).

This value tells us the concentration of the benzene solution. Understanding solution concentration calculations is critical in various fields, including chemistry, biology, and medicine, as it helps in the preparation and analysis of solutions.