   # Which of the following methods would you use to prepare 1.00 L of 0.125 M H 2 SO 4 ? (a) Dilute 20.8 mL of 6.00 M H 2 SO 4 to a volume of 1.00 L. (b) Add 950. mL of water to 50.0 mL of 3.00 M H 2 SO 4 . ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 4, Problem 51PS
Textbook Problem
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## 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 of Solution

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

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