/*! 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 31 Concentrated sulfuric acid ( \(1... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Concentrated sulfuric acid ( \(18.3 M\) ) has a density of \(1.84 \mathrm{~g} / \mathrm{mL}\). (a) How many moles of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) are in each milliliter of solution? (b) What is the mass \(\%\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in the solution?

Short Answer

Expert verified
(a) 0.0183 mol/mL; (b) 97.56%

Step by step solution

01

Find the mass of the solution per milliliter

Given the density of the solution, use the formula: \[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \] Since the density is given as \(1.84 \text{ g/mL}\), for 1 mL of solution, the mass is: \[ \text{Mass} = \text{Density} \times \text{Volume} = 1.84 \text{ g/mL} \times 1 \text{ mL} = 1.84 \text{ g}\]
02

Calculate the number of moles of \(\text{H}_2\text{SO}_4\) in 1 mL of solution

Using the molarity (\(18.3 M\)), calculate the moles: \[ \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \] Rearrange to find moles: \[ \text{Moles} = \text{Molarity} \times \text{Volume in liters} \] Given \(18.3 M\) and \(1 \text{ mL} = 0.001 \text{ L}\), \[ \text{Moles} = 18.3 \text{ M} \times 0.001 \text{ L} = 0.0183 \text{ mol}\]
03

Find the molar mass of \(\text{H}_2\text{SO}_4\)

Calculate the molar mass using the periodic table. The molar mass of \(\text{H}_2\text{SO}_4\) is calculated as follows: \[ \text{Molar mass of } \text{H}_2\text{SO}_4 = 2(1.01) + 32.07 + 4(16.00) = 98.09 \text{ g/mol}\]
04

Calculate the mass of \(\text{H}_2\text{SO}_4\) in 1 mL of solution

Using the number of moles from Step 2 and the molar mass from Step 3: \[ \text{Mass of } \text{H}_2\text{SO}_4 = 0.0183 \text{ mol} \times 98.09 \text{ g/mol} = 1.795 \text{ g}\]
05

Calculate the mass-percent of \(\text{H}_2\text{SO}_4\) in the solution

Using the mass of \(\text{H}_2\text{SO}_4\) found in Step 4 and the mass of the solution from Step 1: \[ \text{Mass percent} = \frac{\text{Mass of } \text{H}_2\text{SO}_4}{\text{Total mass of solution}} \times 100 = \frac{1.795 \text{ g}}{1.84 \text{ g}} \times 100 = 97.56 \text{ }\]

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 91Ó°ÊÓ!

Key Concepts

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

Molarity
Molarity is a term used to express the concentration of a solution. It is defined as the number of moles of solute present in one liter of solution. The formula to calculate molarity is:

\(\text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}}\)

For example, if you have a solution with a molarity of 18.3 M, this means there are 18.3 moles of solute in every liter of that solution. To find the number of moles in a different volume of solution, you can rearrange the formula:

\(\text{Moles of solute} = \text{Molarity} \times \text{Volume in liters}\)

In our exercise, for 1 mL (0.001 L) of solution with a molarity of 18.3 M, the moles of \(\text{H}_2\text{SO}_4\) is calculated as:

\(\text{Moles} = 18.3 \text{ M} \times 0.001 \text{ L} = 0.0183 \text{ mol}\).

Molarity helps in understanding the strength or concentration of a solution, which is crucial in various chemical calculations and reactions.
Density
Density is a measure of how much mass is contained in a given volume. It is an important property in chemistry as it can help determine the mass or volume of substances. The formula for density is:

\(\text{Density} = \frac{\text{Mass}}{\text{Volume}}\)

In our exercise, the density of the sulfuric acid solution is given as 1.84 g/mL. This means that every milliliter of the solution weighs 1.84 grams. Using this relationship, you can find the mass of 1 mL of the solution:

\(\text{Mass} = \text{Density} \times \text{Volume} = 1.84 \text{ g/mL} \times 1 \text{ mL} = 1.84 \text{ g}\)

Understanding density is essential in determining how solutions will behave in different conditions and in performing calculations that involve mass and volume.
Mass Percent
Mass percent, or mass-percent, is a way of expressing the concentration of a component in a mixture as a percentage of the total mass of the mixture. It is calculated using the formula:

\(\text{Mass percent} = \frac{\text{Mass of solute}}{\text{Total mass of solution}} \times 100\)

In the example of our exercise, we first calculated the mass of \(\text{H}_2\text{SO}_4\) in 1 mL of solution to be 1.795 g. Given that the total mass of the solution (1 mL) is 1.84 g, the mass percent is found using the above formula:

\(\text{Mass percent} = \frac{1.795 \text{ g}}{1.84 \text{ g}} \times 100 = 97.56 \text{ }\)

Mass percent is a useful way to express concentrations as it is easy to understand and conveys the proportion of each component in relation to the whole mixture. This concept is particularly valuable in fields like chemistry, pharmaceuticals, and materials science, where precise concentrations need to be managed and communicated.

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

If \(25.98 \mathrm{~mL}\) of \(0.1180 \mathrm{M}\) KOH solution reacts with \(52.50 \mathrm{~mL} .\) of \(\mathrm{CH}_{3} \mathrm{COOH}\) solution, what is the molarity of the acid solution?

Aluminum sulfate, known as cake alum, has a wide range of uses, from dyeing leather and cloth to purifying sewage. In aqueous solution, it reacts with base to form a white precipitate. (a) Write balanced total and net ionic equations for its reaction with aqueous \(\mathrm{NaOH}\). (b) What mass of precipitate forms when \(185.5 \mathrm{~mL}\) of \(0.533 \mathrm{M} \mathrm{NaOH}\) is added to \(627 \mathrm{~mL}\) of a solution that contains \(15.8 \mathrm{~g}\) of aluminum sulfate per liter?

In a car engine, gasoline (represented by \(\mathrm{C}_{8} \mathrm{H}_{18}\) ) does not burn completely, and some CO, a toxic pollutant, forms along with \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). If \(5.0 \%\) of the gasoline forms \(\mathrm{CO}\) : (a) What is the ratio of \(\mathrm{CO}_{2}\) to CO molecules in the exhaust? (b) What is the mass ratio of \(\mathrm{CO}_{2}\) to CO? (c) What percentage of the gasoline must form CO for the mass ratio of \(\mathrm{CO}_{2}\) to \(\mathrm{CO}\) to be exactly \(1 / 1 ?\)

Describe how to determine the oxidation number of sulfur in (a) \(\mathrm{H}_{2} \mathrm{~S}\) and \((\mathrm{b}) \mathrm{SO}_{3}^{2-}\)

Write two sets of equations (both molecular and total ionic) with different reactants that have the same net ionic equation as the following equation: $$ \mathrm{Ba}\left(\mathrm{NO}_{0}\right)_{2}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \longrightarrow \mathrm{BaCO}_{3}(s)+2 \mathrm{NaNO}_{3}(a q) $$

See all solutions

Recommended explanations on Chemistry Textbooks

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.