Q13.57 Calculate the molarity of each a... [FREE SOLUTION]

Chapter 13: Q13.57 (page 548)

Calculate the molarity of each aqueous solution:
0.82 g of ethanol ( $C_{2} H_{5} OH$ ) in 10.5 mL of solution
1.27 g of gaseous ${NH}_{3}$ in 33.5 mL of solution.

Short Answer

Expert verified

The molarity of the 0.82g of table sugar $(C_{2} H_{5} OH)$ in 0.0105L of the solution is 1.7M.
The molarity of the 1.27g of ${NH}_{3}$ in 0.0335L of the solution is 2.23M.

Step by step solution

Concentration of the Solution

The solution can be formed from the dissolution of the solute in the solvent. The solute decides the nature of the solution.

Molarity can be defined as the ratio of the mass of the solute to the volume of the solution present in the litre measure.

$Molarity = \frac{Number 鈥� of 鈥� Moles}{Volume 鈥� of 鈥� the 鈥� Solution 鈥� (鈥� in 鈥� Litre 鈥�)}$

The number of moles can be defined as the ratio of the mass of the atom/molecule and the molar mass of the atom/molecule.

$Number 鈥� of 鈥� Moles = \frac{Mass}{Molar 鈥� Mass}$

Expression to calculate the Concentration

a) Mass of $C_{2} H_{5} OH = 0.82 g$

Molar Mass of $C_{2} H_{5} OH = 46 g / mole$

$Number 鈥� of 鈥� Moles = \frac{Mass}{Molar 鈥� Mass} NumberofMoles = \frac{0.82 g}{46 g / mole} = 0.0178 mole$

localid="1659266499023" $Volume of Solvent = 10.5 mL = 0.0105 L$

$Molarity = \frac{Number 鈥� of 鈥� Moles}{Volume 鈥� of 鈥� the 鈥� Solution (鈥� in 鈥� Litre 鈥�)} Molarity = \frac{0.0178 mole}{0.0105} = 1.7 M$

b) Mass of localid="1659266625860" ${NH}_{3} = 1.27 g$

Molar Mass of ${NH}_{3} = 17 g / mole$

$Number 鈥� of 鈥� Moles = \frac{Mass}{Molar 鈥� Mass} NumberofMoles = \frac{1.27 g}{17 g / mole} = 0.0747 mole$

$Volume of Solvent = 33.5 mL = 0.0335 L$

$Molarity = \frac{Number 鈥� of 鈥� Moles}{Volume 鈥� of 鈥� the 鈥� Solution 鈥� (鈥� in 鈥� Litre 鈥�)} Molarity = \frac{0.0747 mole}{0.0335} = 2.23 M$

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

Calculate the molarity of each aqueous solution:
0.82 g of ethanol ( $C_{2} H_{5} OH$ ) in 10.5 mL of solution
1.27 g of gaseous ${NH}_{3}$ in 33.5 mL of solution.

Short Answer

Step by step solution

Concentration of the Solution

Expression to calculate the Concentration

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Kinetics

Chemical Reactions

Inorganic Chemistry

The Earths Atmosphere

Making Measurements

Study anywhere. Anytime. Across all devices.

91影视

Calculate the molarity of each aqueous solution: 0.82 g of ethanol (C2H5OH) in 10.5 mL of solution 1.27 g of gaseous NH3 in 33.5 mL of solution.

Short Answer

Step by step solution

Concentration of the Solution

Expression to calculate the Concentration

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Kinetics

Chemical Reactions

Inorganic Chemistry

The Earths Atmosphere

Making Measurements

Study anywhere. Anytime. Across all devices.

Calculate the molarity of each aqueous solution:
0.82 g of ethanol ( $C_{2} H_{5} OH$ ) in 10.5 mL of solution
1.27 g of gaseous ${NH}_{3}$ in 33.5 mL of solution.