/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 106 A \(71.2-\mathrm{g}\) sample of ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 106

A \(71.2-\mathrm{g}\) sample of oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), was dissolved in \(1.00\) L of solution. How would you prepare \(1.00 \mathrm{~L}\) of \(0.150 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) from this solution?

Short Answer

Expert verified
Measure 190 mL of the original solution and dilute to 1.00 L to obtain a 0.150 M solution.

Step by step solution

01

Calculate the Moles of Oxalic Acid

First, determine the molar mass of oxalic acid (\(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\)). The formula weight calculation is as follows: H: 1.01 g/mol \(\times 2\), C: 12.01 g/mol \(\times 2\), and O: 16.00 g/mol \(\times 4\). Therefore, the molar mass is \(2 \times 1.01 + 2 \times 12.01 + 4 \times 16.00 = 90.03\) g/mol. Then, use the formula to calculate moles: \( \text{moles} = \frac{71.2 \text{ g}}{90.03 \text{ g/mol}} \approx 0.791 \text{ mol} \) oxalic acid.
02

Calculate the Concentration of the Original Solution

Using the moles calculated previously and the given volume of the solution, determine the original concentration. Since the solution is 1.00 L, the concentration (\(M_1\)) is \( \frac{0.791 \text{ mol}}{1.00 \text{ L}} = 0.791 \text{ M} \). This is the molarity of the original solution.
03

Use Dilution Formula to Find Volume Needed

Use the dilution formula \( M_1V_1 = M_2V_2 \) to find the volume of the original solution required to prepare 1.00 L of 0.150 M solution. Substitute: \( M_1 = 0.791 \text{ M}, M_2 = 0.150 \text{ M}, V_2 = 1.00 \text{ L} \). Solve for \(V_1\): \(0.791 \times V_1 = 0.150 \times 1.00\). Therefore, \(V_1 = \frac{0.150}{0.791} \approx 0.190 \text{ L} \).
04

Prepare the Diluted Solution

Measure \(0.190 \text{ L}\) (or 190 mL) of the original solution and add it to a new container. Then, add distilled water until the total volume reaches 1.00 L. This will give you the desired 0.150 M \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\) solution.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Molarity
Molarity is a way to express the concentration of a solution. It tells us how much solute, the substance being dissolved, is present in a certain volume of solution. The formula to calculate molarity (\( M \)) is:\[M = \frac{\text{moles of solute}}{\text{liters of solution}}\]When working with solutions, this is a crucial concept because it can help us understand how strong or weak a solution is. The more concentrated it is, the higher the molarity. For example, if you have 0.791 moles of oxalic acid dissolved in 1.00 liter, the molarity is 0.791 M. This information is key when you need to perform further calculations or make dilutions.
Dilution Formula
The process of dilution refers to adding more solvent to a solution to decrease its concentration. The dilution formula links the concentrations and volumes of solutions before and after dilution:\[M_1V_1 = M_2V_2\]Where:
  • \( M_1 \) = Initial molarity of the solution
  • \( V_1 \) = Volume of the initial solution
  • \( M_2 \) = Final molarity
  • \( V_2 \) = Final volume
This formula allows us to calculate how much of the initial solution is needed to get a new solution of a desired molarity and volume. For instance, if you needed a 0.150 M solution of oxalic acid in a final volume of 1.00 L, and you started with a 0.791 M solution, the formula gives you the volume of the initial solution required to achieve this.
Oxalic Acid
Oxalic acid, with the chemical formula \( \mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4} \), is a compound you might encounter in chemistry labs or in various natural sources such as plants. It is a dicarboxylic acid, meaning it has two carboxyl groups \((-COOH)\).In the context of preparing solutions, knowing the molar mass of oxalic acid is crucial. It allows us to relate the mass of the compound to the number of moles, which is an important step when calculating molarity.Each molecule of oxalic acid is composed of:
  • 2 Hydrogen atoms
  • 2 Carbon atoms
  • 4 Oxygen atoms
This results in a molar mass of 90.03 g/mol, calculated by adding the atomic masses of all atoms present in one molecule. Understanding this calculation lays the groundwork for using the acid in various chemical experiments and reactions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Determine the oxidation numbers of all the elements in each of the following compounds. (Hint: Look at the ions present.) a. \(\mathrm{Hg}_{2}\left(\mathrm{BrO}_{3}\right)_{2}\) b. \(\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) c. \(\mathrm{CoSeO}_{4}\) d. \(\mathrm{Pb}(\mathrm{OH})_{2}\)

A compound of iron and chlorine is soluble in water. An excess of silver nitrate was added to precipitate the chloride ion as silver chloride. If a \(134.8\) -mg sample of the compound gave \(304.8 \mathrm{mg} \mathrm{AgCl}\), what is the formula of the compound?

A flask contains \(49.8 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) (calcium hydroxide). How many milliliters of \(0.350 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) (sodium carbonate) are required to react completely with the calcium hydroxide in the following reaction? $$ \mathrm{Na}_{2} \mathrm{CO}_{3}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow \mathrm{CaCO}_{3}(s)+2 \mathrm{NaOH}(a q) $$

Gold has compounds containing gold(I) ion or gold(III) ion. A compound of gold and chlorine was treated with a solution of silver nitrate, \(\mathrm{AgNO}_{3}\), to convert the chloride ion in the compound to a precipitate of AgCl. A 162.7-mg sample of the gold compound gave \(100.3 \mathrm{mg} \mathrm{AgCl}\). a. Calculate the percentage of the chlorine in the gold compound. b. Decide whether the formula of the compound is \(\mathrm{AuCl}\) or \(\mathrm{AuCl}_{3}\)

Copper has compounds with copper(I) ion or copper(II) ion. A compound of copper and chlorine was treated with a solution of silver nitrate, \(\mathrm{AgNO}_{3}\), to convert the chloride ion in the compound to a precipitate of \(\mathrm{AgCl}\). A \(59.40-\mathrm{mg}\) sample of the copper compound gave \(86.00 \mathrm{mg}\) AgCl. a. Calculate the percentage of chlorine in the copper compound. b. Decide whether the formula of the compound is \(\mathrm{CuCl}\) or \(\mathrm{CuCl}_{2}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

The Earths Atmosphere

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.