Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 22

The concentrated sulfuric acid we use in the laboratory is 98.0 percent \(\mathrm{H}_{2} \mathrm{SO}_{4}\) by mass. Calculate the molality and molarity of the acid solution. The density of the solution is \(1.83 \mathrm{~g} / \mathrm{mL}\)

Short Answer

Expert verified
The molality of the solution is \(500 m\) and the molarity is \(18.2 M\)

Step by step solution

01

Calculating mass of Sulfuric acid and water

Assume that the total mass of the solution is 100 g. Then, the mass of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is \(98.0\) g, and the mass of water (the solvent) is given by: mass of water = total mass - mass of solute = \(100.0 g - 98.0 g\) = \(2.0 g\)
02

Converting mass to moles

Now convert mass of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) to moles using the molar mass of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (which is \(98.08 g/mol\)): Moles of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) = mass / molar mass = \(98.0 g / 98.08 g/mol\) = \(1.00 mol\)
03

Calculating molality

We can now calculate the molality using the formula: Molality = moles of solute / mass of solvent (in kg), so it gives: Molality = \(1.00 mol / 0.002 kg\) = \(500 mol/ kg\) which is equal to \(500 m\)
04

Calculating the volume of the solution

We need to calculate the volume of the solution to find the molarity. Volume = density / mass = \(1.83 g/mL / 100 g\) = \(0.055 liters\)
05

Calculating the molarity

The formula for molarity is: Molarity = moles of solute / volume of solution (in liters), so: Molarity = \(1.00 mol / 0.055 liters\) = \(18.2 mol/ liter\) which is equal to \(18.2 M\)

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

A solution of \(2.50 \mathrm{~g}\) of a compound having the empirical formula \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{P}\) in \(25.0 \mathrm{~g}\) of benzene is observed to freeze at \(4.3^{\circ} \mathrm{C}\). Calculate the molar mass of the solute and its molecular formula.

Which of the following aqueous solutions has (a) the higher boiling point, (b) the higher freezing point, and (c) the lower vapor pressure: \(0.35 \mathrm{~m}\) \(\mathrm{CaCl}_{2}\) or \(0.90 \mathrm{~m}\) urea? Explain. Assume \(\mathrm{CaCl}_{2}\) to undergo complete dissociation.

Liquids A (molar mass \(100 \mathrm{~g} / \mathrm{mol}\) ) and \(\mathrm{B}\) (molar mass \(110 \mathrm{~g} / \mathrm{mol}\) ) form an ideal solution. At \(55^{\circ} \mathrm{C}, \mathrm{A}\) has a vapor pressure of \(95 \mathrm{mmHg}\) and \(\mathrm{B}\) has a vapor pressure of \(42 \mathrm{mmHg}\). A solution is prepared by mixing equal masses of \(\mathrm{A}\) and \(\mathrm{B}\). (a) Calculate the mole fraction of each component in the solution. (b) Calculate the partial pressures of A and B over the solution at \(55^{\circ} \mathrm{C}\). (c) Suppose that some of the vapor described in (b) is condensed to a liquid in a separate container. Calculate the mole fraction of each component in this liquid and the vapor pressure of each component above this liquid at \(55^{\circ} \mathrm{C}\).

Write the equation representing Raoult's law, and express it in words.

Describe the factors that affect the solubility of a solid in a liquid. What does it mean to say that two liquids are miscible?
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Analysis

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