Problem 52 - Chemistry The Central Science In Si Units

Question

Calculate the number of moles of solute present in each of the following solutions: (a) \(255 \mathrm{~mL}\) of \(1.50 \mathrm{M} \mathrm{HNO}_{3}(a q)\) (b) \(50.0 \mathrm{mg}\) of an aqueous solution that is \(1.50 \mathrm{~m} \mathrm{NaCl}\), (c) \(75.0 \mathrm{~g}\) of an aqueous solution that is \(1.50 \%\) sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) by mass.

Step-by-Step Solution

Answer

(a) 0.3825 moles, (b) 0.000075 moles, (c) 0.00329 moles

Step 1: Calculate Moles for Molarity Solution (a)

To find the moles of solute in a solution with a given molarity, use the formula: \[ \text{moles} = \text{molarity} \times \text{volume in liters} \] For (a), the volume is \(255 \text{ mL} = 0.255 \text{ L}\). Thus, \[ \text{moles of } \mathrm{HNO}_3 = 1.50 \text{ M} \times 0.255 \text{ L} = 0.3825 \text{ moles} \].

Step 2: Calculate Moles for Molality Solution (b)

Molality (\(m\)) is defined as moles of solute per kilogram of solvent, which is water in most cases. Here, we know \(1.50\,m \mathrm{NaCl}\) means 1.50 moles of NaCl per 1 kg of water. \(50.0 \text{ mg}\) of the solution translates to \(0.0500 \text{ g}\). Assuming water density is \(1 \text{ g/mL}\), this is \(0.0000500 \text{ kg}\). Therefore, the moles are calculated as: \[ \text{moles of } \mathrm{NaCl} = 1.50 \times 0.0000500 = 0.000075 \text{ moles} \].

Step 3: Calculate Moles for Mass Percentage Solution (c)

A 1.50% solution by mass means: \(1.50 \text{ g of sucrose per } 100 \text{ g of solution}\). Therefore, \(75.0 \text{ g}\) of solution contains: \[ \text{mass of sucrose} = 75.0 \text{ g} \times \frac{1.50}{100} = 1.125 \text{ g sucrose} \]. The molar mass of sucrose \(\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11}\) is approximately \( 342.30 \text{ g/mol} \). Therefore, \[ \text{moles of sucrose} = \frac{1.125 \text{ g}}{342.30 \text{ g/mol}} = 0.00329 \text{ moles} \]

Key Concepts

Molarity

Molarity is a vital concept in chemistry that describes the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. It's an extremely useful measure for reactions happening in liquid solutions. To calculate molarity, you use the formula:

• \( \text{molarity} (M) = \frac{\text{moles of solute}}{\text{liters of solution}} \).

In practice, if you're given a specific volume of solution and its molarity, you can determine the number of moles of solute present. For instance, in the original exercise, you convert the volume from milliliters to liters, then multiply by the molarity to find the moles. This is essential in solution preparation and stoichiometric calculations, as it allows chemists to precisely determine reactant quantities.

Molality

Molality is another way of expressing concentration, particularly useful when dealing with temperature-dependent environments. Molality (\(m\)) is the number of moles of solute per kilogram of solvent, and it does not change with temperature, providing a reliable measure when temperature shifts are a factor. Here's how you calculate it:

• \( \text{molality} (m) = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \).

In the provided solution, understanding molality allows conversion of small amounts like milligrams to grams, and further to kilograms, facilitating accurate calculations of solute moles per unit mass of solvent. It's particularly valuable in colligative properties and phase diagrams.

Mass Percentage

Mass percentage offers a straightforward means to describe a mixture's composition without concern for volume or temperature. It indicates how much of a mixture's total mass is accounted for by a particular substance. This is given by the formula:

• \( \text{mass percentage} = \left( \frac{\text{mass of solute}}{\text{total mass of solution}} \right) \times 100 \% \).

For solid-liquid solutions, like sucrose in water, the mass percentage is especially useful. If you know the mass percentage and total mass of the solution, you can quickly calculate the solute's mass and then its moles by using the molar mass, as demonstrated in the example. This practical measure is widely used in food, pharmaceuticals, and chemical industries.

Stoichiometry

Stoichiometry is the backbone of chemistry that relates the quantities of reactants and products in a chemical reaction. It's based on the conservation of mass and the premise that reactions proceed in specific ratios, defined by the balanced chemical equation.

• The steps involve using the mole ratios from a balanced equation to convert between the mass or volume of reactants and products, ensuring the reaction is in the right proportions.

In the context of moles, measurements are crucial for determining how much of each substance participates in a reaction. By knowing how to measure moles through molarity, molality, and mass percentage, you can effectively use stoichiometry to predict the outcomes of reactions and to design experiments that maximize efficiency and yield.