Problem 45 - Chemistry: The Central Science

Question

Calculate the molality of each of the following solutions: (a) \(8.66 \mathrm{~g}\) of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) dissolved in \(23.6 \mathrm{~g}\) of carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right)\), (b) \(4.80 \mathrm{~g}\) of \(\mathrm{NaCl}\) dissolved in \(0.350 \mathrm{~L}\) of water.

Step-by-Step Solution

Answer

The short answer to the question is as follows: (a) For the benzene in carbon tetrachloride solution, the molality is calculated to be 4.70 mol/kg. (b) For the sodium chloride in water solution, the molality is calculated to be 0.235 mol/kg.

Step 1: Part (a): Calculation of Molality for Benzene in Carbon Tetrachloride Solution

Step 1: Calculate the molecular weight of benzene (C6H6) One mol of C6H6 has 6 carbon atoms, 6 hydrogen atoms. The molecular weights of carbon and hydrogen are 12.01 g/mol and 1.01 g/mol, respectively. Molecular weight of benzene = (6 × 12.01) + (6 × 1.01) = 78.12 g/mol Step 2: Calculate the moles of benzene Given, 8.66 g of benzene is dissolved. Now, we will calculate the moles of benzene. Moles of benzene = mass of benzene / molecular weight of benzene = 8.66 g / 78.12 g/mol = 0.1109 mol Step 3: Convert the mass of carbon tetrachloride to kilograms Given mass of carbon tetrachloride = 23.6 g To convert g to kg, divide by 1000 Mass of carbon tetrachloride = 23.6 / 1000 = 0.0236 kg Step 4: Calculate the molality of the solution Molality of solution = moles of benzene / mass of carbon tetrachloride in kg = 0.1109 mol / 0.0236 kg = 4.70 mol/kg (molality)

Step 2: Part (b): Calculation of Molality for Sodium Chloride in Water Solution

Step 1: Calculate the molecular weight of sodium chloride (NaCl) One mol of NaCl has 1 sodium atom & 1 chlorine atom. The molecular weights of sodium and chlorine are 22.99 g/mol and 35.45 g/mol, respectively. Molecular weight of NaCl = 22.99 + 35.45 = 58.44 g/mol Step 2: Calculate the moles of sodium chloride Given, 4.80 g of sodium chloride is dissolved. Now, we will calculate the moles of sodium chloride. Moles of sodium chloride = mass of sodium chloride / molecular weight of sodium chloride = 4.80 g / 58.44 g/mol = 0.0822 mol Step 3: Convert the mass of water to kilograms Given volume of water = 0.350 L Density of water = 1.00 g/mL = 1000 g/L. Now, we will calculate the mass of water (in g). Mass of water = volume × density = 0.350 L × 1000 g/L = 350 g To convert g to kg, divide by 1000 Mass of water = 350 / 1000 = 0.350 kg Step 4: Calculate the molality of the solution Molality of solution = moles of sodium chloride / mass of water in kg = 0.0822 mol / 0.350 kg = 0.235 mol/kg (molality)

Key Concepts

Solution Concentration

Understanding solution concentration is crucial when studying chemistry. Concentration refers to the amount of solute that is dissolved in a solvent. The solute is the substance that is dissolved, while the solvent is the substance that does the dissolving, often a liquid. There are various ways to express the concentration of a solution, including molarity, molality, and percentage composition.

Molality, specifically, is a measure of the concentration of a solution expressed as the number of moles of solute per kilogram of solvent. Unlike molarity, which uses the volume of the solution, molality uses the mass of the solvent. This makes molality particularly useful when the solution's temperature changes, as it does not vary with temperature like molarity does because volume can expand or contract with temperature.

In the provided example, the molality is calculated for two different solutions. A precise determination of the solute's mass and the solvent's mass in kilograms is vital. From there, molality can be calculated using the formula:\[\begin{equation}Molality (m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}\end{equation}\]This calculation is fundamental in various scientific fields, including chemistry, biology, and environmental science. It helps predict how the solution will behave under different conditions, which is essential in laboratory and industrial settings.

Molar Mass

The molar mass of a substance defines the mass of one mole of that substance. It is a fundamental concept in chemistry because it links the mass of a substance to the quantity of particles or moles.

The molar mass is usually expressed in grams per mole (\(g/mol\)) and can be calculated by summing the atomic masses of all the atoms in a molecule of the substance. This value is critical when converting between moles and grams of a substance. For example, the molar mass of benzene (\(C_6H_6\)), as demonstrated in the solution, is calculated by adding the molar masses of six carbon atoms and six hydrogen atoms to get\(78.12g/mol\).

Knowing the molar mass allows us to calculate the number of moles in a given mass of a substance using the formula:\[\begin{equation}Moles = \frac{\text{mass}}{\text{molar mass}}\end{equation}\]This step is essential when preparing solutions with precise concentrations, or when performing stoichiometric calculations in chemical reactions.

Moles Calculation

Calculating moles is a critical skill in chemistry that requires understanding molar mass. A mole represents a specific number of particles, specifically Avogadro's number (\(6.022 \times 10^{23}\) particles), whether they are atoms, ions, or molecules.

For substance quantity conversions, chemists use the formula mentioned before to find the number of moles given the substance's mass and molar mass. It's a three-step process involving the accurate measurement of the sample mass, knowledge of the molar mass, and application of the conversion formula.
In the exercise solutions we looked at, the number of moles of benzene and sodium chloride were calculated. Correctly calculating the number of moles is fundamental for further steps in the molality calculation and provides insight into the quantity of the substance involved in a reaction or a solution on a molecular level.

The ability to calculate moles aids in understanding reaction yields, solution concentrations, and molecular interactions. It's a cornerstone for topics like stoichiometry, thermodynamics, and kinetics, and should be confidently understood by students and professionals working in the sciences.