/*! 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 19 Calculate the number of grams in... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 19

Calculate the number of grams in \(500 \mathrm{~mL}\) of each of the following solutions: (a) \(0.100 M\) \(\mathrm{Na}_{2} \mathrm{SO}_{4},\) (b) \(0.250 \mathrm{M} \mathrm{Fe}\left(\mathrm{NH}_{4}\right)_{2}\left(\mathrm{SO}_{4}\right)_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O},\) (c) \(0.667 \mathrm{M} \mathrm{Ca}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{ON}\right)_{2}\)

Short Answer

Expert verified
(a) 7.10 g, (b) 49.02 g, (c) 123.64 g.

Step by step solution

01

Understand the Problem

We need to calculate the mass in grams of solute present in 500 mL of three different molar solutions: (a) \(0.100\,M\,\mathrm{Na}_2\mathrm{SO}_4\), (b) \(0.250\,M\,\mathrm{Fe}\left(\mathrm{NH}_4\right)_2\left(\mathrm{SO}_4\right)_2\cdot 6\mathrm{H}_2\mathrm{O}\), and (c) \(0.667\,M\,\mathrm{Ca}\left(\mathrm{C}_9\mathrm{H}_6\mathrm{ON}\right)_2\). We do this by using the molar mass of each compound and the formula: \( \text{mass (g)} = \text{molarity (M)} \times \text{volume (L)} \times \text{molar mass (g/mol)} \).
02

Identify Relevant Information for (a)

For \(\mathrm{Na}_2\mathrm{SO}_4\), the molar mass is calculated as follows:- \(2 \times \text{Na} = 2 \times 22.99 = 45.98\) g/mol- \(1 \times \text{S} = 32.07\) g/mol- \(4 \times \text{O} = 4 \times 16.00 = 64.00\) g/molThus, \(\text{Molar mass of } \mathrm{Na}_2\mathrm{SO}_4 = 45.98 + 32.07 + 64.00 = 142.05\) g/mol.
03

Calculate Mass for 0.100 M \(\mathrm{Na}_2\mathrm{SO}_4\)

Use the formula: \( \text{mass (g)} = 0.100\,\text{M} \times 0.500\,\text{L} \times 142.05\,\text{g/mol} = 7.10\,\text{g}\).
04

Identify Relevant Information for (b)

For \(\mathrm{Fe}\left(\mathrm{NH}_4\right)_2\left(\mathrm{SO}_4\right)_2\cdot 6\mathrm{H}_2\mathrm{O}\), calculate its molar mass:- \(1 \times \text{Fe} = 55.85\) g/mol- \(2 \times \text{NH}_4 = 2 \times 18.04 = 36.08\) g/mol- \(2 \times \text{SO}_4 = 2 \times 96.07 = 192.14\) g/mol- \(6 \times \text{H}_2\text{O} = 6 \times 18.02 = 108.12\) g/molThus, \(\text{Molar mass} = 55.85 + 36.08 + 192.14 + 108.12 = 392.19\) g/mol.
05

Calculate Mass for 0.250 M \(\mathrm{Fe}\left(\mathrm{NH}_4\right)_2\left(\mathrm{SO}_4\right)_2\cdot 6\mathrm{H}_2\mathrm{O}\)

Use the formula: \( \text{mass (g)} = 0.250\,\text{M} \times 0.500\,\text{L} \times 392.19\,\text{g/mol} = 49.02\,\text{g}\).
06

Identify Relevant Information for (c)

For \(\mathrm{Ca}\left(\mathrm{C}_9\mathrm{H}_6\mathrm{ON}\right)_2\), calculate its molar mass:- \(1 \times \text{Ca} = 40.08\) g/mol- \(2 \times \text{C}_9\text{H}_6\text{ON} = 2 \times (9 \times 12.01 + 6 \times 1.01 + 16.00 + 14.01) = 2 \times 165.18 = 330.36\) g/molThus, \(\text{Molar mass} = 40.08 + 330.36 = 370.44\) g/mol.
07

Calculate Mass for 0.667 M \(\mathrm{Ca}\left(\mathrm{C}_9\mathrm{H}_6\mathrm{ON}\right)_2\)

Use the formula: \( \text{mass (g)} = 0.667\,\text{M} \times 0.500\,\text{L} \times 370.44\,\text{g/mol} = 123.64\,\text{g}\).

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.

Understanding Solution Concentration
Solution concentration is a crucial concept in chemistry that describes how much solute is present in a given volume of solution. It's usually expressed in terms of molarity (M), which is the number of moles of solute per liter of solution. To find the concentration:
  • Calculate the moles of the solute: Use the molar mass of the solute to convert grams to moles.
  • Determine the volume of the solution in liters.
  • Divide the number of moles of solute by the volume in liters to find the molarity.
This concentration measure tells us about the amount of dissolved substance in the solution, which is crucial for predicting how the solution will behave in chemical reactions.
Demystifying Stoichiometry
Stoichiometry is the part of chemistry that involves calculating the quantities of reactants and products in chemical reactions. It's like a map that guides us on how much of a substance we'll need or how much we can expect to get:
  • Use balanced chemical equations: They show the ratio of moles of reactants and products.
  • Calculate moles needed: From the given molarity and volume (in liters), you can determine the moles of the reactants or products.
  • Convert moles to grams: Use the molar mass of a compound to convert from moles to grams, helping us measure amounts more practically.
By using stoichiometry, we relate quantities in a reaction and ensure all substances are accounted for, maintaining the law of conservation of mass.
Mastering Chemistry Calculations
Chemistry calculations often involve multiple steps and concepts, but breaking them down can make them more manageable. For tasks like calculating the mass of solutes, follow this guide:
  • Find the molar mass: Sum the atomic masses of each element present in one molecule of the compound.
  • Use the concentration formula: \( \text{mass (g)} = \text{molarity (M)} \times \text{volume (L)} \times \text{molar mass (g/mol)} \)
  • Convert volumes if needed: Remember that 500 mL is the same as 0.5 L.
Performing these calculations accurately allows chemists to predict the outcomes of reactions and prepare solutions with the desired concentrations easily. It's all about applying the right formula at the correct step to obtain meaningful results.

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 urine specimen has a chloride concentration of \(150 \mathrm{meq} / \mathrm{L}\). If we assume that the chloride is present in urine as sodium chloride, what is the concentration of \(\mathrm{NaCl}\) in \(\mathrm{g} / \mathrm{L} ?\)

Calculate the molar concentrations of \(1.00-\mathrm{mg} / \mathrm{L}\) solutions of each of the following. (a) \(\mathrm{AgNO}_{3},\) (e) \(\mathrm{HCl},\) (f) \(\mathrm{HClO}_{4}\). (b) \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3},\) (c) \(\mathrm{CO}_{2},\) (d) \(\left(\mathrm{NH}_{4}\right)_{4} \mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\)

The sulfur content of an iron pyrite ore sample is determined by converting it to \(\mathrm{H}_{2} \mathrm{~S}\) gas, absorbing the \(\mathrm{H}_{2} \mathrm{~S}\) in \(10.0 \mathrm{~mL}\) of \(0.00500 \mathrm{M} \mathrm{I}_{2},\) and then back-titrating the excess \(\mathrm{I}_{2}\) with \(0.00200 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} .\) If \(2.6 \mathrm{~mL} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) is required for the titration, how many milligrams of sulfur are contained in the sample? Reactions: $$ \begin{aligned} \mathrm{H}_{2} \mathrm{~S}+\mathrm{I}_{2} \rightarrow \mathrm{S}+2 \mathrm{I}^{-}+2 \mathrm{H}^{+} \\ \mathrm{I}_{2}+2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-} \rightarrow 2 \mathrm{I}^{-}+\mathrm{S}_{4} \mathrm{O}_{6}^{2-} \end{aligned} $$

Define the term "equivalent weight," used for electrolytes in clinical chemistry. Why is this used?

Calculate the molar concentrations of all the cations and anions in a solution prepared by mixing \(10.0 \mathrm{~mL}\) each of the following solutions: \(0.100 \mathrm{M} \mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{2}, 0.100 M \mathrm{KNO}_{3},\) and \(0.100 M \mathrm{~K}, \mathrm{SO}_{4}\) \end{aligned}
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Making Measurements

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.