Question

Given the molarity and density for each of the following acidic solutions, calculate the mass/mass percent concentration: (a) \(6.00 \mathrm{MHCl}(d=1.10 \mathrm{~g} / \mathrm{mL})\) (b) \(1.00 \mathrm{M} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(d=1.01 \mathrm{~g} / \mathrm{mL})\) (c) \(0.500 \mathrm{M} \mathrm{HNO}_{3}(d=1.01 \mathrm{~g} / \mathrm{mL})\) (d) \(3.00 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}(d=1.18 \mathrm{~g} / \mathrm{mL})\)

Short answer

The mass percents are (a) 19.89%, (b) 5.94%, (c) 3.12%, (d) 24.94%.

Step-by-step solution

Understanding Mass Percent Formula

The mass percent of a solution is given by the formula \( \text{Mass \, Percent} = \left( \frac{\text{mass of solute}}{\text{mass of solution}} \right) \times 100 \). In each part of this exercise, we need to calculate the mass of the solute and the total mass of the solution.

Calculate for Part (a) HCl Solution

For 1 liter of \(6.00 \text{ M HCl}\), "6.00 M" means there are \(6.00 \, \text{mol}\) of HCl per liter. The molar mass of HCl is approximately \(36.46 \, \text{g/mol}\). The mass of HCl in 1 liter is \( 6.00 \, \text{mol} \times 36.46 \, \text{g/mol} = 218.76 \, \text{g} \). Given \( d = 1.10 \, \text{g/mL} \), the mass of 1 liter of solution is \( 1000 \, \text{mL} \times 1.10 \, \text{g/mL} = 1100 \, \text{g} \). Thus, the mass percent is \( \left( \frac{218.76}{1100} \right) \times 100 \approx 19.89\% \).

Calculate for Part (b) HC2H3O2 Solution

For 1 liter of \(1.00 \text{ M} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\), you have \(1.00 \, \text{mol}\) of acetic acid. The molar mass is \(60.05 \, \text{g/mol}\). So the mass of acetic acid is \( 1.00 \, \text{mol} \times 60.05 \, \text{g/mol} = 60.05 \, \text{g} \). Given \( d = 1.01 \, \text{g/mL} \), the mass of 1 liter of solution is \( 1000 \, \text{mL} \times 1.01 \, \text{g/mL} = 1010 \, \text{g} \). Therefore, the mass percent is \( \left( \frac{60.05}{1010} \right) \times 100 \approx 5.94\% \).

Calculate for Part (c) HNO3 Solution

For 1 liter of \(0.500 \, \text{M} \mathrm{HNO}_{3}\), you have \(0.500 \, \text{mol}\) of nitric acid. The molar mass is \(63.01 \, \text{g/mol}\). The mass of \(\mathrm{HNO}_{3}\) is \( 0.500 \, \text{mol} \times 63.01 \, \text{g/mol} = 31.505 \, \text{g} \). Given \( d = 1.01 \, \text{g/mL} \), the mass of 1 liter of solution is \( 1000 \, \text{mL} \times 1.01 \, \text{g/mL} = 1010 \, \text{g} \). So, the mass percent is \( \left( \frac{31.505}{1010} \right) \times 100 \approx 3.12\% \).

Calculate for Part (d) H2SO4 Solution

For 1 liter of \(3.00 \text{ M} \mathrm{H}_{2} \mathrm{SO}_{4}\), you have \(3.00 \, \text{mol}\) of sulfuric acid. The molar mass is \(98.08 \, \text{g/mol}\). The mass of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is \( 3.00 \, \text{mol} \times 98.08 \, \text{g/mol} = 294.24 \, \text{g} \). With \( d = 1.18 \, \text{g/mL} \), the mass of 1 liter of solution is \( 1000 \, \text{mL} \times 1.18 \, \text{g/mL} = 1180 \, \text{g} \). Thus, the mass percent is \( \left( \frac{294.24}{1180} \right) \times 100 \approx 24.94\% \).

Key concepts

Molarity

Molarity is a way of expressing the concentration of a solution. It defines the number of moles of a solute in one liter of solution.
In mathematical terms, it's often written as: \[ ext{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}. \]For instance, if we have a 6.00 M HCl solution, it means there are 6.00 moles of hydrochloric acid in every liter of the solution.
The idea of molarity is really useful in chemical reactions because it lets us know how many molecules of a substance are present. This ensures experiments and reactions can be accurately conducted.
Molarity is commonly used in various areas, including chemistry labs and industrial applications, to precisely prepare solutions.

Density

Density is another critical measurement in understanding solutions. It reflects how much a substance weighs relative to its volume, usually expressed in grams per milliliter (g/mL).
When dealing with solutions, density helps determine the total mass of a given volume of the solution. This mass is crucial for calculating concentrations like mass percent concentration.
For example, if a solution of HCl has a density of 1.10 g/mL, it indicates that 1 mL of this solution weighs 1.10 grams.
Understanding the density of a solution can also be essential for identifying the nature of the solution, comparing substances, or determining the composition.

Chemical Solutions

In the world of chemistry, solutions are homogeneous mixtures where the solute is uniformly distributed within the solvent. The solute is the substance that's dissolved, while the solvent is the substance in which it is dissolved.
Chemical solutions can be found everywhere, from saltwater to acids like hydrochloric acid (HCl).
Key characteristics of chemical solutions include:

• Uniformity: Solutions have a single-phase appearance, with no visible difference between their components.
• Composition: A solution's properties depend directly on the concentration of its solute, which defines its chemical behavior.
• Varied Uses: Solutions play crucial roles in experiments, manufacturing, and everyday products.

These solutions can range from simple ones used in a laboratory to complex mixes in industrial applications.

Acidic Solutions

Acidic solutions are those that have higher concentrations of hydrogen ions (H⁺). They tend to have a pH less than 7.
Different acids, such as hydrochloric acid (HCl), acetic acid (HC₂H₃O₂), nitric acid (HNO₃), and sulfuric acid (H₂SO₄), are often used in various solutions. Their molarity and density help us calculate their concentration in solutions.
Some common properties of acidic solutions include:

• Reactivity: Acids often react with metals, bases, and other substances.
• Sour Taste: Commonly perceived in foods like citrus fruits due to the presence of acids.
• Conductivity: Acids generally conduct electricity due to the presence of ions.

Acidic solutions are utilized in numerous fields, such as cleaning agents, food preservation, and chemical manufacturing.
Understanding their molarity, density, and chemical interaction is crucial for safely and effectively using these solutions.