BuyFindarrow_forward

Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Solutions

Chapter
Section
BuyFindarrow_forward

Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

Which of the following methods would you use to prepare 1.00 L of 0.125 M H2SO4?

(a) Dilute 20.8 mL of 6.00 M H2SO4 to a volume of 1.00 L.

(b) Add 950. mL of water to 50.0 mL of 3.00 M H2SO4.

Interpretation Introduction

Interpretation:

It should be determined that the correct method to prepare 1.00L of 0.125MH2SO4 from the given methods.

Concept introduction:

  • Concentration of solutions can be expressed in various terms; molarity is one such concentration expressing term.
  • Molarity (M) of a solution is the number of gram moles of a solute present in one liter of the solution.

     Molarity=MassperlitreMolecular massMolarity=W×1000m×V

Where W, V and m are the mass of solute, volume of solution and molar mass of the solute respectively.

  W=m×V×M1000

  • A solution containing one gram mole or 0.1 gram of solute per litre of solution is called molar solution. A solution containing one gram mole or 0.1 gram of solute per litre of solution is called molar solution.
  • Concentrationofsubstance=Amountof substancevolumeofthesubstance
  • Amountof substance=Concentrationofsubstance×volumeofthesubstance
Explanation

In method a 20.8mL of 6.00MH2SO4 diluted to a volume of 1.00L

Amount of H2SO4 in the 20.8mL solution can be calculated as follows,

  AmountofH2SO4 = CH2SO4×VH2SO4 =6.0 molH2SO4L×20.8×10-3L = 0.1248molH2SO4

0.1248molH2SO4 is contained in the new volume of 1.0L of diluted solution.

Therefore the final concentration of the diluted solution is,

ConcentrationofH2SO4,CH2SO4 = [H2SO4] = 0.1248molH2SO41L =0

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started
Sect-4.8 P-4.11CYUSect-4.8 P-4.12CYUSect-4.8 P-4.13CYUSect-4.9 P-4.14CYUSect-4.9 P-1.1ACPSect-4.9 P-2.1ACPSect-4.9 P-3.1ACPSect-4.9 P-4.1ACPSect-4.9 P-4.2ACPSect-4.9 P-4.3ACPSect-4.9 P-4.4ACPCh-4 P-1PSCh-4 P-2PSCh-4 P-3PSCh-4 P-4PSCh-4 P-5PSCh-4 P-6PSCh-4 P-7PSCh-4 P-8PSCh-4 P-9PSCh-4 P-10PSCh-4 P-11PSCh-4 P-12PSCh-4 P-13PSCh-4 P-14PSCh-4 P-15PSCh-4 P-16PSCh-4 P-17PSCh-4 P-18PSCh-4 P-19PSCh-4 P-20PSCh-4 P-21PSCh-4 P-22PSCh-4 P-23PSCh-4 P-24PSCh-4 P-25PSCh-4 P-26PSCh-4 P-27PSCh-4 P-28PSCh-4 P-29PSCh-4 P-30PSCh-4 P-31PSCh-4 P-32PSCh-4 P-33PSCh-4 P-34PSCh-4 P-35PSCh-4 P-36PSCh-4 P-37PSCh-4 P-38PSCh-4 P-39PSCh-4 P-40PSCh-4 P-41PSCh-4 P-42PSCh-4 P-43PSCh-4 P-44PSCh-4 P-45PSCh-4 P-46PSCh-4 P-47PSCh-4 P-48PSCh-4 P-49PSCh-4 P-50PSCh-4 P-51PSCh-4 P-52PSCh-4 P-53PSCh-4 P-54PSCh-4 P-55PSCh-4 P-56PSCh-4 P-57PSCh-4 P-58PSCh-4 P-59PSCh-4 P-60PSCh-4 P-61PSCh-4 P-62PSCh-4 P-63PSCh-4 P-64PSCh-4 P-65PSCh-4 P-66PSCh-4 P-67PSCh-4 P-68PSCh-4 P-69PSCh-4 P-70PSCh-4 P-71PSCh-4 P-72PSCh-4 P-73PSCh-4 P-74PSCh-4 P-75PSCh-4 P-76PSCh-4 P-77PSCh-4 P-79GQCh-4 P-80GQCh-4 P-81GQCh-4 P-82GQCh-4 P-83GQCh-4 P-84GQCh-4 P-85GQCh-4 P-86GQCh-4 P-87GQCh-4 P-88GQCh-4 P-89GQCh-4 P-90GQCh-4 P-91GQCh-4 P-92GQCh-4 P-93GQCh-4 P-94GQCh-4 P-95GQCh-4 P-96GQCh-4 P-97GQCh-4 P-98GQCh-4 P-99GQCh-4 P-100GQCh-4 P-101GQCh-4 P-102GQCh-4 P-103GQCh-4 P-104GQCh-4 P-105GQCh-4 P-106GQCh-4 P-107GQCh-4 P-108GQCh-4 P-109GQCh-4 P-110GQCh-4 P-111GQCh-4 P-112GQCh-4 P-113GQCh-4 P-115GQCh-4 P-116GQCh-4 P-117GQCh-4 P-118GQCh-4 P-119GQCh-4 P-120GQCh-4 P-121ILCh-4 P-122ILCh-4 P-123ILCh-4 P-125ILCh-4 P-126ILCh-4 P-127ILCh-4 P-128ILCh-4 P-129ILCh-4 P-130ILCh-4 P-135SCQCh-4 P-136SCQCh-4 P-137SCQCh-4 P-139SCQCh-4 P-141SCQCh-4 P-142SCQ

Additional Science Solutions

Find more solutions based on key concepts

Show solutions add

Hypoglycemia as a disease is relatively common. T F

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List)

What causes an aging giant star to produce a planetary nebula?

Horizons: Exploring the Universe (MindTap Course List)

A 10.0-kg monkey climbs a uniform ladder with weight 1.20 102 N and length L = 3.00 m as shown in Figure P12.1...

Physics for Scientists and Engineers, Technology Update (No access codes included)