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Chapter 14: Problem 117

How many moles of the indicated solute does each of the following solutions contain? a. 1.5 L of \(3.0 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution b. 35 mL of 5.4 M NaCl solution c. 5.2 L of \(18 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution d. \(0.050 \mathrm{L}\) of \(1.1 \times 10^{-3} \mathrm{M}\) NaF solution

Short Answer

Expert verified
a. 4.5 moles of H鈧係O鈧� b. 0.189 moles of NaCl c. 93.6 moles of H鈧係O鈧� d. \(5.5 \times 10^{-5}\) moles of NaF

Step by step solution

01

Identify the Volume and Concentration

We are given the volume, 1.5 L, and the concentration, 3.0 M (moles/liter), of the H鈧係O鈧� solution.
02

Calculate the Moles

Use the formula: Moles = Volume 脳 Concentration. Moles = \(1.5 L \times 3.0 \frac{moles}{L} = 4.5 moles\) There are 4.5 moles of H鈧係O鈧� in the solution. #b. 35 mL of 5.4 M NaCl solution#
03

Identify the Volume and Concentration

We are given the volume, 35 mL, and the concentration, 5.4 M, of the NaCl solution. First, we need to convert the volume to liters. Volume = \(\frac{35 mL}{1000} = 0.035 L\)
04

Calculate the Moles

Use the formula: Moles = Volume 脳 Concentration. Moles = \(0.035 L \times 5.4 \frac{moles}{L} = 0.189 moles\) There are 0.189 moles of NaCl in the solution. #c. 5.2 L of 18 M H鈧係O鈧� solution#
05

Identify the Volume and Concentration

We are given the volume, 5.2 L, and the concentration, 18 M (moles/liter), of the H鈧係O鈧� solution.
06

Calculate the Moles

Use the formula: Moles = Volume 脳 Concentration. Moles = \(5.2 L \times 18 \frac{moles}{L} = 93.6 moles\) There are 93.6 moles of H鈧係O鈧� in the solution. #d. 0.050 L of 1.1 脳 10鈦宦� M NaF solution#
07

Identify the Volume and Concentration

We are given the volume, 0.050 L, and the concentration, \(1.1 \times 10^{-3}\) M (moles/liter), of the NaF solution.
08

Calculate the Moles

Use the formula: Moles = Volume 脳 Concentration. Moles = \(0.050 L \times 1.1 \times 10^{-3} \frac{moles}{L} = 5.5 \times 10^{-5} moles\) There are \(5.5 \times 10^{-5}\) moles of NaF in the solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Concentration
Concentration refers to the amount of a substance present in a certain volume of a solution. It's vital to understand this concept when dealing with chemical solutions, as it tells us how "strong" or "weak" a solution is. In chemistry, the most common measure of concentration is molarity. Molarity (M) is defined as the number of moles of solute per liter of solution. It's expressed as moles per liter (mol/L).

For example, a solution with a molarity of 3.0 M has 3 moles of the solute in every liter of the solution. Knowing the molarity helps chemists understand how much chemical substance they have in a certain volume of liquid, which is crucial for reactions.
Volume to Moles Conversion
To find out how many moles of a substance are in a solution, you need to convert the volume of the solution to moles. This process involves using the concept of molarity.
  • First, make sure the volume is in liters. Remember, 1000 mL = 1 L.
  • Then, multiply the volume (in liters) by the concentration (molarity) of the solution.
The formula you'll use is:\[ \text{Moles} = \text{Volume (L)} \times \text{Concentration (M)} \]By using this method, you can easily convert the known volume of a solution into the amount of substance in terms of moles.
Insights Into Chemical Solutions
A chemical solution is a homogeneous mixture composed of two or more substances. In a solution, the substance in larger quantity is usually the solvent, while the one in smaller quantity is the solute. Solutions are transparent and can have varying degrees of concentration, governed by the proportion of solute to solvent.

Understanding the characteristics of solutions and how to manipulate their concentration is essential in many fields, such as pharmaceuticals, environmental science, and industrial chemistry. These skills help chemists to create the desired reaction conditions and to manage concentrations effectively.
Tackling Educational Chemistry Problems
Educational chemistry problems often involve calculations related to concentration and moles. These problems are designed to cultivate critical thinking and problem-solving skills in students. When facing such problems, it鈥檚 important to break down each step clearly:
  • Identify the known variables: volume, concentration, etc.
  • Convert units where necessary to maintain consistency.
  • Apply the relevant formulas (like volume to moles conversion).
  • Perform calculations with care, double-checking each step.
These exercises help students to gain familiarity with real-world applications of chemistry principles. They are a key part of learning about chemical reactions and processes.

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Most popular questions from this chapter

A solution that contains 1 equivalent of acid or base per liter is said to be a _______ solution.

Standard solutions of calcium ion used to test for water hardness are prepared by dissolving pure calcium carbonate, \(\mathrm{CaCO}_{3}\), in dilute hydrochloric acid. A 1.745 -g sample of \(\mathrm{CaCO}_{3}\) is placed in a 250.0 -mL volumetric flask and dissolved in HCl. Then the solution is diluted to the calibration mark of the volumetric flask. Calculate the resulting molarity of calcium ion.

For convenience, one form of sodium hydroxide that is sold commercially is the saturated solution. This solution is \(19.4 \mathrm{M},\) which is approximately \(50 \%\) by mass sodium hydroxide. What volume of this solution would be needed to prepare 3.50 L of \(3.00 M\) NaOH solution?

Suppose \(50.0 \mathrm{mL}\) of \(0.250 \mathrm{M} \mathrm{CoCl}_{2}\) solution is added to \(25.0 \mathrm{mL}\) of \(0.350 \mathrm{M} \mathrm{NiCl}_{2}\) solution. Calculate the concentration, in moles per liter, of each of the ions present after mixing. Assume that the volumes are additive.

What volume of \(1.00 \mathrm{M} \mathrm{NaOH}\) is required to neutralize each of the following solutions? a. \(25.0 \mathrm{mL}\) of \(0.154 \mathrm{M}\) acetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) b. \(35.0 \mathrm{mL}\) of \(0.102 \mathrm{M}\) hydrofluoric acid, \(\mathrm{HF}\) c. \(10.0 \mathrm{mL}\) of \(0.143 \mathrm{M}\) phosphoric acid, \(\mathrm{H}_{3} \mathrm{PO}_{4}\) d. \(35.0 \mathrm{mL}\) of \(0.220 \mathrm{M}\) sulfuric acid, \(\mathrm{H}_{2} \mathrm{SO}_{4}\)
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