/*! 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 104 A 230 -mL sample of a \(0.275-M ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 104

A 230 -mL sample of a \(0.275-M \mathrm{CaCl}_{2}\) solution is left on a hot plate overnight; the following morning, the solution is 1.10\(M .\) What volume of water evaporated from the 0.275\(M \mathrm{CaCl}_{2}\) solution?

Short Answer

Expert verified
172.5 mL of water evaporated from the 0.275 M CaCl鈧� solution.

Step by step solution

01

Calculate the initial moles of CaCl鈧�

First, we need to find the initial moles of CaCl鈧� using the initial concentration and volume of the solution. The formula for calculating moles is: Moles = Concentration 脳 Volume Here, the initial concentration of CaCl鈧� is 0.275 M, and the initial volume is 230 mL. We will convert the volume to liters: 230 mL 脳 (1 L / 1000 mL) = 0.230 L Now, we can calculate the initial moles of CaCl鈧�: Moles of CaCl鈧� = 0.275 M 脳 0.230 L = 0.06325 mol
02

Calculate the final moles of CaCl鈧�

Since the moles of CaCl鈧� remain constant after the evaporation of water, the final moles of CaCl鈧� are the same as the initial moles: Final moles of CaCl鈧� = 0.06325 mol
03

Calculate the final volume of the solution

Next, we will find the final volume of the CaCl鈧� solution using the final concentration and the final moles of CaCl鈧�. The formula to calculate the volume is: Volume = Moles / Concentration The final concentration of CaCl鈧� is given as 1.10 M. We can now calculate the final volume of the solution: Final volume = 0.06325 mol / 1.10 M = 0.0575 L
04

Calculate the volume of water that evaporated

Now, we can calculate the volume of water that evaporated from the initial solution by comparing the initial and final volumes of the solution: Initial volume - Final volume = Volume of water evaporated 0.230 L - 0.0575 L = 0.1725 L
05

Convert the volume to milliliters

Finally, we will convert the volume of evaporated water to milliliters: 0.1725 L 脳 (1000 mL / 1 L) = 172.5 mL So, 172.5 mL of water evaporated from the 0.275 M CaCl鈧� 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 indicates how many moles of a solute are present in one liter of solution. Understanding molarity is crucial in chemistry, especially when preparing or diluting solutions. Let's break down the key points:
  • Molarity (\( M \)) is determined using the formula: \( M = \frac{\text{moles of solute}}{\text{liters of solution}} \).
  • This measurement allows you to easily calculate how much of a substance is in a given quantity of water or solvent."
  • In the original exercise, the molarity of the calcium chloride (\( \text{CaCl}_2 \)) solution changes from \( 0.275 M \) to \( 1.10 M \) due to evaporation.
Molarity helps quickly determine how the removal of the solvent affects the concentration of the solution. It depicts why solutions become more concentrated as they lose a solvent, such as water.
Volume Change
Volume change occurs when the amount of solvent in a solution changes, while the amount of solute remains the same. This is a crucial concept when studying the effects of evaporation on a solution:
  • Initially, we have a solution with a specific volume 鈥� in this case, 230 mL.
  • After the water evaporates, the volume of the solution decreases, even though the solute amount remains unchanged.
  • This decrease in volume results in an increase in solution concentration.
In our exercise, as the water evaporates, the total volume reduces significantly, causing the solution's concentration to increase from \( 0.275 M \) to \( 1.10 M \). Understanding volume change is essential to predict how the solution's concentration adjusts over time due to evaporation or other processes.
Evaporation
Evaporation is a process where liquid turns into vapor without boiling. It's a natural way for a liquid phase to lose solvent molecules, leading to a concentrated solution. Here's how evaporation impacts a solution:
  • The amount of solute does not alter, but the solvent quantity decreases.
  • This loss of water molecules results in a smaller volume and a higher concentration of the solution.
  • In our problem, leaving the \( \text{CaCl}_2 \) solution on the hot plate overnight led to the evaporation of 172.5 mL of water.
Evaporation is common in everyday scenarios where water bodies lose moisture to the atmosphere. Understanding its effect helps in practical and laboratory settings where precise concentrations are necessary.
Calculation of Moles
Calculating moles is an essential component in solution chemistry. Moles are a standard unit for measuring the amount of substance. Here's a simple guide:
  • To find the number of moles, multiply the concentration by the volume (in liters) of the solution.
  • Use the formula: \( \text{Moles} = \text{Concentration} \times \text{Volume} \).
  • In the exercise, the initial moles of \( \text{CaCl}_2 \) were found using \( 0.275 M \times 0.230 L = 0.06325 \ ext{mol} \).
Accurate calculation of moles is crucial. It ensures that chemical reactions achieve the desired outcomes, as moles determine the proportions in which substances react. Recognizing how initial and final moles remain constant, independent of evaporation, underpins many solution-based problems in chemistry.

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

Assign oxidation numbers to all the atoms in each of the following. a. \(\mathrm{SrCr}_{2} \mathrm{O}_{7} \quad\) g. \(\mathrm{PbSO}_{3}\) b. \(\mathrm{CuCl}_{2} \quad \quad\) h. \(\mathrm{PbO}_{2}\) c. \(\mathrm{O}_{2} \quad\quad\quad\) i. \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) d. \(\mathrm{H}_{2} \mathrm{O}_{2} \quad\quad \mathrm{j} . \mathrm{CO}_{2}\) e. \(\mathrm{MgCO}_{3} \quad\) k. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{3}\) f. \(\mathrm{Ag} \quad\quad\quad \)l. \(\mathrm{Cr}_{2} \mathrm{O}_{3}\)

One of the classic methods for determining the manganese content in steel involves converting all the manganese to the deeply colored permanganate ion and then measuring the absorption of light. The steel is first dissolved in nitric acid, producing the manganese(II) ion and nitrogen dioxide gas. This solution is then reacted with an acidic solution containing the periodate ion; the products are the permanganate and iodate ions. Write balanced chemical equations for both of these steps.

An average human being has about 5.0 \(\mathrm{L}\) of blood in his or her body. If an average person were to eat 32.0 \(\mathrm{g}\) of sugar (sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}, 342.30 \mathrm{g} / \mathrm{mol}\) ), and all that sugar were dissolved into the bloodstream, how would the molarity of the blood sugar change?

A 25.00-mL sample of hydrochloric acid solution requires 24.16 mL of 0.106 M sodium hydroxide for complete neutralization. What is the concentration of the original hydrochloric acid solution?

Acetylsalicylic acid is the active ingredient in aspirin. It took 35.17 mL of 0.5065 M sodium hydroxide to react completely with 3.210 g of acetylsalicylic acid. Acetylsalicylic acid has one acidic hydrogen. What is the molar mass of acetylsalicylic acid?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

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.