Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 138

What volume of \(0.8 \mathrm{M}-\mathrm{AlCl}_{3}\) solution should be mixed with \(50 \mathrm{ml}\) of \(0.2 \mathrm{M}-\mathrm{CaCl}_{2}\) solution to get a solution of chloride ion concentration equal to \(0.6 \mathrm{M} ?\) (a) \(5.56 \mathrm{ml}\) (c) \(50 \mathrm{ml}\)

Short Answer

Expert verified
Answer: Approximately 1.56 ml.

Step by step solution

01

Calculate the moles of chloride ions in the \(\mathrm{CaCl}_{2}\) solution

First, we need to determine the moles of chloride ions present in the given volume of \(0.2 \mathrm{M}-\mathrm{CaCl}_{2}\) solution. To do this, we multiply the volume of the solution by the molarity of chloride ions. Note that 1 mole of \(\mathrm{CaCl}_{2}\) has 2 moles of chloride ions: Moles of chloride ions = Volume × Molarity × 2 Moles of chloride ions = 50 ml × 0.2 M × 2 Moles of chloride ions = 0.02 L × 0.2 mol/L × 2 Moles of chloride ions = 0.004 mol
02

Determine the required total moles of chloride ions in the final solution

We need to find how many moles of chloride ions are required in total to achieve the desired concentration of \(0.6 \mathrm{M}\) in the final solution. For that, we'll first calculate the final volume of the solution. Let's denote the volume of the \(\mathrm{AlCl}_{3}\) solution as \(V\). The final volume of the solution will be \((50 + V) \mathrm{ml}\). Total moles of chloride ions = Final concentration × Final volume Total moles of chloride ions = 0.6 M × (0.05 L + V)
03

Calculate the moles of chloride ions contributed by \(\mathrm{AlCl}_{3}\) solution

Now, we'll determine the moles of chloride ions contributed by the \(\mathrm{AlCl}_{3}\) solution. Note that 1 mole of \(\mathrm{AlCl}_{3}\) has 3 moles of chloride ions: Moles of chloride ions from \(\mathrm{AlCl}_{3}\) = Volume × Molarity × 3 Moles of chloride ions from \(\mathrm{AlCl}_{3}\) = V × 0.8 M × 3
04

Apply the principle of conservation of mass for chloride ions

The total moles of chloride ions in the final solution should be equal to the sum of the moles of chloride ions contributed by the \(\mathrm{CaCl}_{2}\) and \(\mathrm{AlCl}_{3}\) solutions. 0.6 M × (0.05 L + V) = 0.004 mol + (V × 0.8 M × 3)
05

Solve for the volume of \(\mathrm{AlCl}_{3}\) solution needed

Rearrange the equation from Step 4 to solve for the volume V. V × 2.4 M - 0.6 M × V = 0.004 mol - 0.0012 L V × 1.8 M = 0.004 mol - 0.0012 L V = (0.004 mol - 0.0012 L) / 1.8 M V = 0.001556 L
06

Convert the volume to milliliters and choose the correct answer

Finally, we'll convert the volume of \(\mathrm{AlCl}_{3}\) solution needed from liters to milliliters: V = 0.001556 L × 1000 ml/L V ≈ 1.56 ml This value tells us that we need to mix approximately \(1.56 \mathrm{ml}\) of \(0.8 \mathrm{M}-\mathrm{AlCl}_{3}\) solution with \(50 \mathrm{ml}\) of \(0.2 \mathrm{M}-\mathrm{CaCl}_{2}\) solution to achieve a chloride ion concentration equal to \(0.6 \mathrm{M}\). None of the given answer choices match this value, which means that the provided choices are incorrect.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

The percentage of \(\mathrm{Fe}(\mathrm{III})\) present in iron ore \(\mathrm{Fe}_{0.93} \mathrm{O}_{1.00}\) is \((\mathrm{Fe}=56)\) (a) 94 (b) 6 (c) \(21.5\) (d) 15

A quantity of \(0.2 \mathrm{~g}\) of an organic compound containing, \(\mathrm{C}, \mathrm{H}\) and \(\mathrm{O}\), on combustion yielded \(0.147 \mathrm{~g} \mathrm{CO}_{2}\) and \(0.12 \mathrm{~g}\) water. The percentage of oxygen in it is (a) \(73.29 \%\) (b) \(78.45 \%\) (c) \(83.23 \%\) (d) \(89.50 \%\)

One molecule of haemoglobin will combine with four molecules of oxygen. If \(1.0 \mathrm{~g}\) of haemoglobin combines with \(1.642 \mathrm{ml}\) of oxygen at body temperature \(\left(27^{\circ} \mathrm{C}\right)\) and a pressure of 760 torr, what is the molar mass of haemoglobin? (a) \(6,00,000\) (b) \(1,50,000\) (c) 15,000 (d) 60,000

A mixture of methane and ethylene in the ratio of a:b by volume occupies \(30 \mathrm{ml}\). On complete combustion, the mixture yield \(40 \mathrm{ml}\) of \(\mathrm{CO}_{2} .\) What volume of \(\mathrm{CO}_{2}\) would have been obtained if the ratio would have been b:a? (a) \(50 \mathrm{ml}\) (b) \(30 \mathrm{ml}\) (c) \(40 \mathrm{ml}\) (d) \(60 \mathrm{ml}\)

Molecular mass of dry air is (a) less than moist air (b) greater than moist air (c) equal to moist air (d) may be greater or less than moist air
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

Kinetics

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.

Sign-up for free