Problem 67 - Chemistry: The Central Science

Question

Calculate (a) the number of grams of solute in \(0.250 \mathrm{~L}\) of \(0.175 \mathrm{M} \mathrm{KBr},(\mathrm{b})\) the molar concentration of a solution containing \(14.75 \mathrm{~g}\) of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) in \(1.375 \mathrm{~L},(\mathrm{c})\) the volume of \(1.50 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) in milliliters that contains \(2.50 \mathrm{~g}\) of solute.

Step-by-Step Solution

Answer

The short answers for each part are: (a) There are 5.21 grams of solute in 0.250 L of 0.175 M KBr solution. (b) The molar concentration of a solution containing 14.75 g of Ca(NO3)2 in 1.375 L is 0.0654 M. (c) The volume of 1.50 M Na3PO4 solution containing 2.50 g of solute is 10.163 mL.

Step 1: a) Grams of solute in 0.250 L of 0.175 M KBr

To find the grams of solute, we first need to find the number of moles of solute. We can do this by using the formula: moles of solute = volume of solution (in L) × molar concentration (M) Next, we will need to find the molecular weight of KBr and then the weight of the solute using the formula: grams of solute = moles of solute × molecular weight of solute The molecular weight of KBr is the sum of the molecular weight of K and Br. The molecular weight of K is 39 g/mol and for Br, it is 80 g/mol. Molecular weight of KBr = 39 g/mol + 80 g/mol = 119 g/mol Now, we can find the number of moles of solute. moles of solute = 0.250 L × 0.175 M = 0.04375 mol Now we can find the grams of solute. grams of solute = 0.04375 mol × 119 g/mol = 5.20625 g So, there are 5.20625 grams of solute in 0.250 L of 0.175 M KBr solution.

Step 2: b) Molar concentration of a solution containing 14.75 g of Ca(NO3)2 in 1.375 L

To find the molar concentration of the solution, we first need to find the moles of solute from the given weight. We can do this by using the formula: moles of solute = grams of solute / molecular weight of solute Next, we will need to find the molecular weight of Ca(NO3)2. Molecular weight of Ca(NO3)2 = molecular weight of Ca + 2 × molecular weight of N + 6 × molecular weight of O Using the periodic table, the molecular weights are: Ca = 40 g/mol, N = 14 g/mol, and O = 16 g/mol. Molecular weight of Ca(NO3)2 = 40 g/mol + 2 × 14 g/mol + 6 × 16 g/mol = 164 g/mol Now, we can find the moles of solute. moles of solute = 14.75 g / 164 g/mol = 0.089939 g/mol Now, we can find the molar concentration of the solution. Molar Concentration = moles of solute / volume of solution (in L) = 0.089939 g/mol / 1.375 L = 0.065394 M So, the molar concentration of a solution containing 14.75 g of Ca(NO3)2 in 1.375 L is 0.065394 M.

Step 3: c) Volume of 1.50 M Na3PO4 in milliliters that contains 2.50 g of solute

To find the volume of the solution, we first need to find the moles of solute from the given weight. We can do this by using the formula: moles of solute = grams of solute / molecular weight of solute Next, we will need to find the molecular weight of Na3PO4. Molecular weight of Na3PO4 = 3 × molecular weight of Na + molecular weight of P + 4 × molecular weight of O Using the periodic table, the molecular weights are: Na = 23 g/mol, P = 31 g/mol, and O = 16 g/mol. Molecular weight of Na3PO4 = 3 × 23 g/mol + 31 g/mol + 4 × 16 g/mol = 164 g/mol Now, we can find the moles of solute. moles of solute = 2.50 g / 164 g/mol = 0.015244 g/mol Now, we can find the volume of the solution using the molar concentration. Volume of solution (in L) = moles of solute / molar concentration = 0.015244 g/mol / 1.5 M = 0.010163 L Now, we can convert the volume from liters to milliliters. Volume of solution (in mL) = 0.010163 L × 1000 mL/L = 10.163 mL So, the volume of 1.50 M Na3PO4 solution containing 2.50 g of solute is 10.163 mL.

Key Concepts

Molar Mass

Understanding the molar mass of a substance is crucial in chemistry, particularly when it comes to stoichiometry and solution preparation. The molar mass is defined as the mass of one mole of a particular substance and is expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all the atoms in the molecule as provided by the periodic table. For example, the molar mass of potassium bromide (KBr) is calculated by adding the atomic mass of potassium (K), which is 39 g/mol, to that of bromine (Br), which is 80 g/mol, yielding a molar mass of 119 g/mol.

With this information, we can determine how much of the substance we have in a given amount of solution, allowing us to convert between moles and grams. This step is essential in solving stoichiometry problems, as shown in the provided exercise.

Stoichiometry

Stoichiometry is the section of chemistry that pertains to the quantitative relationships between the reactants and products in a chemical reaction. It is based on the mole concept and the conservation of mass. Understanding stoichiometry is essential for predicting the amounts of substances consumed and produced in reactions. In practical terms, it allows us to use balanced equations to calculate the mass of reactants needed to produce a certain amount of product, or vice versa.

In the exercise, stoichiometry is applied to calculate the grams of solute in a solution and the volume of solution needed to contain a certain amount of solute. The process involves converting volumes and concentrations to moles and then using molar mass to move between moles and grams of substances.

Solution Concentration

Solution concentration is a measure of the quantity of solute that is contained in a given volume of solution. It is often expressed in molarity (M), which is the number of moles of solute per liter of solution. Understanding how to calculate solution concentration allows one to prepare solutions with desired chemical properties for various applications.

In the problems provided, concentration plays a key role in determining both the mass of solute and the volume of the solution required. The second part of the exercise demonstrates how to calculate the molar concentration by dividing the moles of solute by the volume of the solution. This concept is pivotal because it bridges the quantitative gap between the amount of solute and the volume in which it is dissolved.

Mole Concept

The mole concept is a fundamental cornerstone in chemistry that describes the amount of substance. One mole is defined as the amount of a substance that contains the same number of entities (atoms, molecules, ions, etc.) as atoms in 12 grams of carbon-12. This number is Avogadro’s number, which is approximately 6.022×10²³ entities per mole. The mole concept is integral for calculating the number of particles, relating masses to each other, and converting between grams and moles.

In the provided problems, the mole concept is used to determine the number of moles from the given mass of a solute and then using this information to calculate the molar concentration or to find the volume of the solution containing a known mass of solute. This concept shows that at the core of understanding concentrations and reactions is the ability to correctly count and convert particles into tangible quantities we can measure.