Problem 59 - Chemistry The Molecular Science

Question

Calculate the mass fraction and the weight percent of the solute in each of these solutions: (a) \(14.0 \mathrm{~g} \mathrm{~K}_{2} \mathrm{CrO}_{4}\) in \(225 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\). (b) \(4.56 \mathrm{~g}\) ethanol in \(50.0 \mathrm{~g}\) benzene. (c) \(15.0 \mathrm{~g}\) methanol in \(89.0 \mathrm{~g}\) ethanol. (d) \(14.5 \mathrm{~mL}\) ethylene glycol \((d=1.11 \mathrm{~g} / \mathrm{mL})\) in \(200 . \mathrm{g} \mathrm{H}_{2} \mathrm{O}\).

Step-by-Step Solution

Answer

(a) 5.86%; (b) 8.36%; (c) 14.42%; (d) 7.45% weight percent.

Step 1: Understand the problem

The task is to find two values for each solution: the mass fraction and the weight percent of the solute in the solution. The mass fraction is the ratio of the mass of the solute to the total mass of the solution, and the weight percent is the mass fraction multiplied by 100.

Step 2: Solve for part (a)

Solution (a) contains \(14.0 \text{ g K}_2\text{CrO}_4\) and \(225 \text{ g H}_2\text{O}\). Calculate the total mass of the solution by adding these two masses: \(14.0 + 225 = 239.0 \text{ g}\). The mass fraction is \( \frac{14.0}{239.0} \approx 0.0586\). The weight percent is \(0.0586 \times 100 \approx 5.86\%\).

Step 3: Solve for part (b)

In solution (b), there are \(4.56 \text{ g ethanol}\) and \(50.0 \text{ g benzene}\). Total mass of the solution is \(4.56 + 50.0 = 54.56 \text{ g}\). The mass fraction is \( \frac{4.56}{54.56} \approx 0.0836\). The weight percent is \(0.0836 \times 100 \approx 8.36\%\).

Step 4: Solve for part (c)

In solution (c), \(15.0 \text{ g methanol}\) and \(89.0 \text{ g ethanol}\) are present. The total mass is \(15.0 + 89.0 = 104.0 \text{ g}\). The mass fraction is \( \frac{15.0}{104.0} \approx 0.1442\). The weight percent is \(0.1442 \times 100 \approx 14.42\%\).

Step 5: Solve for part (d)

In solution (d), \(14.5 \text{ mL ethylene glycol}\) is used with a density of \(1.11 \text{ g/mL}\). First, convert the volume of ethylene glycol to mass: \(14.5 \times 1.11 = 16.095 \text{ g}\). The total mass of the solution is \(16.095 + 200 = 216.095 \text{ g}\). The mass fraction is \( \frac{16.095}{216.095} \approx 0.0745\). The weight percent is \(0.0745 \times 100 \approx 7.45\%\).

Key Concepts

Weight Percent

Weight percent is a way to express the concentration of a solute in a solution. It's calculated by finding the mass fraction and then multiplying by 100. This results in a percentage that tells us how much of the solute makes up the solution's total mass.
For example, if you have 5 grams of solute in a 100 grams solution, the weight percent is dictated by \[\text{Weight Percent} = \left(\frac{5}{100}\right) \times 100 = 5\% \]
This means that 5% of the solution's weight comes from the solute. In daily contexts, weight percent helps when mixing solutions in chemistry labs or industries to ensure correct proportions.

Solution Chemistry

In solution chemistry, solutions are homogeneous mixtures made of two or more substances. The two main components are the solute, which is the substance being dissolved, and the solvent, the medium in which the solute is dissolved.
Understanding solution chemistry is crucial for various applications such as:

• Creating accurate chemical formulations
• Performing chemical reactions efficiently
• Ensuring solubility and reactions proceed correctly

For instance, when table salt (sodium chloride) is dissolved in water, salt becomes the solute, and water is the solvent. Solution properties like conductivity and pH can change based on the solute's nature and concentration.

Mass Calculation

Mass calculation is a key technique to determine how much a solute weighs in a solution, as well as the overall weight of a solution. This part of chemistry ensures precise measurements for accurate experimental outcomes.
For instance, if you're given a concentration scenario where a solute like potassium chromate is dissolved in water, calculating the total mass involves adding the mass of both the solute and the solvent.Here's the basic step:

• Add the mass of the solute to the mass of the solvent.

An example is:\[\text{Total Mass} = \text{Mass of Solute} + \text{Mass of Solvent}\]
For solution preparation, mass calculation helps achieve a desired solute concentration.

Chemical Solutions

Chemical solutions are prevalent in both natural and industrial processes, where a solute is uniformly dispersed within a solvent. These solutions can vary in terms of concentration and components based on the use and requirement.
Chemical solutions are crucial because they are the medium for:

• Nutrient transport within living organisms
• Pharmaceutical preparations
• Reactions environments in labs and industries

For example, in a sugar-water solution, sugar acts as the solute and dissolves evenly in water, the solvent. Properties like boiling and freezing points of these solutions often differ from pure substances, affecting how they are used in different scenarios.