   # In a coffee-cup calorimeter, 1.60 g NH 4 NO 3 is mixed with 75.0 g water at an initial temperature of 25.00°C. After dissolution of the salt, the final temperature of the calorimeter contents is 23.34°C. Assuming the solution has a heat capacity of 4.18 J / °C · g and assuming no heat loss to the calorimeter, calculate the enthalpy change for the dissolution of NH 4 NO 3 in units of kJ/mol. ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 7, Problem 62E
Textbook Problem
1 views

## In a coffee-cup calorimeter, 1.60 g NH4NO3 is mixed with 75.0 g water at an initial temperature of 25.00°C. After dissolution of the salt, the final temperature of the calorimeter contents is 23.34°C. Assuming the solution has a heat capacity of 4.18 J/°C · g and assuming no heat loss to the calorimeter, calculate the enthalpy change for the dissolution of NH4NO3 in units of kJ/mol.

Interpretation Introduction

Interpretation: The enthalpy change for dissolving the NH4NO3 salt in Water has to be calculated.

Concept Introduction: The heat capacity C is defined as the ratio of heat absorbed and temperature change. It can be given by,

C = Heatabsorbed(inJ)Temperature(in°C)

• The required heat of a one gram of substance raise to its temperature by one degree celsius is called specific heat capacity.

Absorbed heat (J) = Specific heat capacity ×Temperature change(°C)×mass ofsubstance (g) (1)

For in above equation heat is:

q= S×M×T (2)

S is specific heat capacity (J°C.g)

M is mass of sample ( g )

T is temperature change ( °C )

For the process no heat loss to the surroundings means then the heat is

(absorbed) =-q (released) (3)

### Explanation of Solution

Explanation

Given data:

Water weight = 75g

Mass of NH4NO3 = 1.60g

Initial temperature of water = 25°C

Final temperature of solution =23.34°C

Specific heat capacity of the solution = 4.18 JCg-1

The final weight of solution is 75g +1.60g = 76.60g

To determine: Heat lost by solution

Heat lost by solution is equal to heat gained by NH4NO3

Heat lost by solution = 4.18×76.60×(25-23.34)

= 532 J

ΔH is in J/g units is = 532J1

### 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

Find more solutions based on key concepts
Taking supplements of fish oil is recommended for those who don't like fish. T F

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

Air (a diatomic ideal gas) at 27.0C and atmospheric pressure is drawn into a bicycle pump that has a cylinder w...

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

Whats a suspension feeder?

Oceanography: An Invitation To Marine Science, Loose-leaf Versin 