Question

Concentrated aqueous ammonia has a molarity of 14.8 and a density of \(0.90 \mathrm{g} / \mathrm{cm}^{3} .\) What is the molality of the solution? Calculate the mole fraction and weight percent of \(\mathrm{NH}_{3}.\)

Short answer

Molality: 22.83 mol/kg; Mole fraction NH3: 0.291; Weight percent NH3: 27.98%.}

Step-by-step solution

Determine Mass of NH3

To find the mass of ammonia in the solution, first calculate the volume of 1 liter of solution, which is equivalent to 1000 mL or 1000 cm extsuperscript{3}. Using the density of 0.90 g/cm extsuperscript{3}, the mass of 1 liter of the solution is 0.90 g/cm extsuperscript{3} \( \times \) 1000 cm extsuperscript{3} = 900 g. Given the molarity of 14.8 M, calculate the number of moles of NH extsubscript{3} in 1 liter: 14.8 moles. Then, find the mass using the molar mass of NH extsubscript{3}, which is 17.03 g/mol. So, the mass of NH extsubscript{3} is 14.8 moles \( \times \) 17.03 g/mol = 251.84 g.

Calculate Mass of Water

Subtract the mass of NH extsubscript{3} from the total mass of the solution to find the mass of water. Thus, the mass of water is 900 g (total mass of solution) - 251.84 g (mass of NH extsubscript{3}) = 648.16 g.

Determine Molality

Molality is defined as the moles of solute per kilogram of solvent. We have 14.8 moles of NH extsubscript{3} and the mass of water is 648.16 g, equivalent to 0.64816 kg. Thus, the molality \( (m) = \frac{14.8 ext{ moles}}{0.64816 ext{ kg}} = 22.83 ext{ mol/kg} \).

Calculate Mole Fraction of NH3

The mole fraction is the ratio of the moles of NH extsubscript{3} to the total moles (NH extsubscript{3} + H extsubscript{2}O). Calculate moles of water: \( \frac{648.16 \text{ g}}{18.015 \text{ g/mol}} \approx 36.0 \text{ moles} \). Therefore, the total moles = 14.8 moles + 36.0 moles = 50.8 moles. Thus, the mole fraction \( (X_{NH_3}) = \frac{14.8}{50.8} = 0.291 \).

Calculate Weight Percent of NH3

The weight percent is the mass percent of NH extsubscript{3} in the solution. We already calculated the mass of NH extsubscript{3} as 251.84 g and the total mass of the solution as 900 g. So, the weight percent \( = \frac{251.84 \text{ g}}{900 \text{ g}} \times 100\% \approx 27.98\% \).

Key concepts

Molarity

Molarity is a fundamental concept in chemistry that is used to express the concentration of a solution. It is defined as the number of moles of solute per liter of solution. For example, when we talk about a 14.8 M aqueous ammonia solution, it means there are 14.8 moles of ammonia (\(NH_3\)) in every liter of the solution. This allows chemists to easily communicate and work with chemical concentrations in a quantitative way.

Molarity is particularly useful because it directly relates to the volume of the solution, which is often what's used in the lab. However, it can vary with temperature, as increasing temperature can expand the liquid, thus altering the volume. Therefore, while convenient, molarity needs to be used with awareness of its dependency on temperature.

Mole Fraction

Mole fraction is another way to express concentration and is particularly important in mixtures. It denotes the ratio of the moles of a component to the total moles in the mixture. It is dimensionless, providing a simple way to compare quantities in a mixture irrespective of volume changes or temperature fluctuations.

To find the mole fraction of ammonia in our example, you add the moles of ammonia and the moles of water, then divide the moles of ammonia by this total. With 14.8 moles of ammonia and approximately 36.0 moles of water, the mole fraction of ammonia becomes 0.291. This means that 29.1% of the particles in the solution are ammonia molecules. Note that mole fractions always add up to 1 and are useful in vapour pressure calculations and colligative properties.

Weight Percent

Weight percent provides another perspective on concentration by expressing the proportion of the solute's mass relative to the total mass of the solution. It is calculated by dividing the mass of the solute by the total mass of the solution and then multiplying by 100 to get a percentage.

For the ammonia solution, we calculated a weight percent of approximately 27.98%. This means that in this solution, about 28% of its weight is due to ammonia. Weight percent is helpful in fields like material science and food chemistry where mass rather than volume is of primary interest.

Due to the ease of measuring mass changes (like evaporation), weight percent gives a reliable measure for formulations that don't involve notable temperature or pressure changes.

Density of Solution

Density is a crucial property that relates mass to volume, often expressed in grams per cubic centimeter (g/cm³). It helps in converting between mass and volume, making it vital for understanding solution concentrations like molarity.

In the problem, the density of the solution is specified as 0.90 g/cm³, indicating that each cubic centimeter of the solution weighs 0.90 grams. This information is essential when we need to calculate the total mass of any given volume of the solution, such as finding out how much one liter (1000 cm³) weighs. This was found to be 900 grams in our exercise.

Understanding the density allows chemists to transition between amounts of solute and solution for various calculations. In laboratory settings, precise density measurements can influence experimental outcomes, especially in quantitative analyses.