Problem 33 - General Chemistry: Principles and Modern Applications

Question

What are the molarities of the following solutes? (a) sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) if \(150.0 \mathrm{g}\) is dissolved per \(250.0 \mathrm{mL}\) of water solution (b) urea, \(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2},\) if \(98.3 \mathrm{mg}\) of the \(97.9 \%\) pure solid is dissolved in \(5.00 \mathrm{mL}\) of aqueous solution (c) methanol, \(\mathrm{CH}_{3} \mathrm{OH},(d=0.792 \mathrm{g} / \mathrm{mL})\) if \(125.0 \mathrm{mL}\) is dissolved in enough water to make 15.0 L of solution

Step-by-Step Solution

Answer

The molarities for the solutes are; (a) Sucrose solution: 2.08 M. (b) Urea solution: 3.13 M. (c) Methanol solution: 0.206 M.

Step 1: Compute Molarity of Sucrose Solution

Molar mass of sucrose \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) is \(342.3 \mathrm{g/mol}\). First, convert the given mass of sucrose (150.0 g) to moles using the molar mass. Secondly, the volume of water (250.0 mL) should be converted to liters, since molarity is defined in terms of liters of solution. Lastly, use the formula for molarity which is \(\text{M} = \frac{\text{moles of solute}}{\text{liters of solution}}\).

Step 2: Compute Molarity of Urea Solution

The molar mass of urea, \(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}\), is \(60.06 \mathrm{g/mol}\). Given mass (98.3 mg) needs to be first converted to grams. Then, to take into account the purity, multiply the mass by 0.979. Convert this mass to moles using the molar mass. Convert the volume of water (5.00 mL) to liters. Finally, determine the molarity by dividing moles of solute by liters of solution.

Step 3: Compute Molarity of Methanol Solution

The molar mass of methanol, \(\mathrm{CH}_{3} \mathrm{OH}\), is \(32.04 \mathrm{g/mol}\). With given volume (125.0 mL) and the density (0.792 g/mL), calculate the mass of methanol in grams. Convert this mass to moles. Then, use the total solution volume (15.0 L) to determine the molarity.

Key Concepts

Chemical Solutions

Chemical solutions are homogeneous mixtures composed of two or more substances. The component present in the largest amount is typically known as the solvent, and the other component(s) are solutes.
In aqueous solutions, water acts as the solvent, while the solute is the substance dissolved in water.
The concentration of a chemical solution refers to how much solute is present compared to the amount of solvent.
One common way to express concentration is the molarity (M), which is the number of moles of solute per liter of solution.
This makes it a crucial concept in chemistry for understanding reaction stoichiometry and chemical equilibria.

Molar Mass Calculation

Molar mass is an important concept that helps us relate the mass of a substance to the amount in moles. It is defined as the mass of one mole of a given substance, and is expressed in grams per mole (g/mol).
To calculate the molar mass of a compound, add up the atomic masses of all atoms present in one molecule of that compound. These atomic masses can be found on the periodic table.
Understanding molar mass allows chemists to measure out a precise amount of a substance for reactions or to determine the concentration of solutions. For example:

• Sucrose \( ext{C}_{12} ext{H}_{22} ext{O}_{11}\) - The molar mass is 342.3 g/mol, calculated by adding up the atomic masses of 12 carbon, 22 hydrogen, and 11 oxygen atoms.
• Urea \(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}\) - The molar mass is 60.06 g/mol by summing the masses of carbon, oxygen, and nitrogen atoms.
• Methanol \(\mathrm{CH}_{3}\mathrm{OH}\) - Molar mass is 32.04 g/mol obtained by adding the masses of carbon, hydrogen, and oxygen atoms.

Molarity Formula

Molarity is a fundamental concept in chemistry that describes the concentration of a solute in a solution. It is defined by the formula:\[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \]
To use this formula effectively, it is important to carefully carry out the following steps:

• First, calculate the number of moles of the solute using its mass and molar mass. This involves dividing the mass of the solute (in grams) by its molar mass (in g/mol).
• Second, convert the solution volume from milliliters to liters, because molarity is expressed in terms of liters. Remember that 1000 mL equals 1 L.
• Lastly, apply the molarity formula by dividing the moles of solute by the liters of solution, providing the concentration in molarity (M).

Understanding this process helps in various applications such as preparing solutions with desired concentrations and conducting quantitative analyses in laboratory settings.