rinciples of General Chemistry 4/18/17, 11:28 8edb5e3b0@placeholder. 18112 edu.

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rinciples of General Chemistry 4/18/17, 11:28 8edb5e3b0@placeholder. 18112 edu. Printing is for personal, private use only. No part of this book may be reproduced or blisher's prior permission. Violators will be prosecuted. Enthalpy (Heat) of solutims for the Dissolutiom Sal On Name of salt Trial 1 Trial 2 1. Mass of salt (g) 2. Moles of salt (mol) Mass of calorimeter (g) 4. Mass of calorimeter water (g) 5. Mass of water Cg) 23.9°C 24.2 C Initial temperature of water 7. Final temperature of mixture from graph 8. Instructor's approval of gaph Caleulatlons for Enthalpy (Hea) ofsolution for the Dissolution ora salt 1. Temperature change of solution, ATcC 2. Heat change of water U) 3. change of salt ()(Obuain its specific heat from Table 25.1). Heat 5. AH, (thmol sal), equation 25.12 6. Average AH Whol salt) Optional. See instructor. significant figures. Show calculations for Trial 1. Report the result with the comect number of

Explanation / Answer

To solve the questions at all we should know the salt we have, since we don´t know it, I will show you the procedure you have to follow when you know the salt.

1. The diference of temperature would be:

Trial 1 = 46.6 -24.2 = 22.4 C

Trial 2 = 47.2 - 23.9 = 23.3 C

2. The heat change of water:

You have the mass and you are asking for the energy transferred, you need to multiply the mass by the specific heat of water.

Specific Heat of water is 4.184 J/ g C

For trial 1 : 18.851 g * 4.184 J/g C * 22.4 C = 1,766.74 J

For trial 2: 20.0188 g * 4.184 J/g C * 23.3C = 1951.576 J

3. Heat change of salt.

To calculate these value you need to know the salt you have, this is because every salt has its own specific heat, to calculate this value you must use the next equation similar to the one for the water:

Heat change of salt: mass of salt * Specific heat of salt (J / g C) * Temperature Change.

For academic purpose, let´s suppose that the salt we have is Lithium Chloride, LiCl, from the table we know it has an specific heat of 1.13 J / gC

For trial 1 : 5.006 g * 1.13 J/gC * 22.4 C = 126.71 J

For trial 2: 5.0631 * 1.13 * 23.3 = 133.30 J

4. The total enthalpy change is just the sum of enthalpies from the water and salt.

Total enthalpy change is = Enthalpy of water + Enthalpy of salt

If we continue with the example we would have:

For trial 1 = 1766.74 J + 126.71 J = 1893.45 J

For trial 2 = 1951.576 J + 133.3 J = 2084.876 J

5. The enthalpy solution J/mol, for this part you have to know the salt you have in order to calculte the number of moles:

Let´s continue the example with the LiCl

We have to calculate number of moles, to do this we need the molecular weight of LiCl

Molecular weight of Li is 6.941

molecular weight of Cl is 35.5

Molecular weight of salt is 42.441

Moles for trial 1 = 5.006 / 42.4 = 0.118 moles

Moles for trial 2 = 5.0631 / 42.4 = 0.119 moles

The Enthalpy of dissolution would be

For trial 1= 1893.45 / 0.118 = 16 046.186 J /mol

For trial 2 = 2084.876 / 0.119 = 17 520 J/ mol

The average would be:

(16 046.186 + 17520) / 2 = 16 783.076 J/mol

This last operations were made as examples, you need to ask your profesor wich is the salt you have, then you have to replace the numbers.

I must add that 2 assumptions were made to do this calculations:

1. The energy transferred to the calorimeter is negligible (equal to zero)

2. The change of temperature of salt is the same change of temperature of water. This is that the initial temperature of water is the same initial temperature of salt.

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