So you've got to do two things.
1) Perform the investigation
2) Calculate the molar heat of solution
and...this is what I did.
PRACTICAL
AIM
To determine the heat energy adsorbed or lost from the two compounds Ammonium Chloride and Sodium Hydroxide
(Hmmm, the compounds that you're using might not be the same....)
(For this practical we used: ∆H=-mC∆T)
EQUIPMENT
1 x 100ml Cylinder (To measure the water the amount of water)
2 x 35mL Beakers (To place the compounds in)
5.35g of Ammonium Chloride (We actually calculated the mass
)
4g of Sodium Hydroxide (Same as Ammonium Cl)
2 Paper Cups (This is where the process took place. Water + Compound)
Microscaler (Very accurate scaling measure. We measured the mass of the compounds through this)
Thermometer (Measure the change in temperature)
100ml of water (The amount of water to be put with the compound)
CALCULATING THE MASS
So we basically calculated the mass of the compounds. This is what our teacher told us.
"We want 0.1n"
In other words, we want 0.1 moles of the substance.
In order to calculate the mass we use the formula :
n = m/mm
Number of Moles = Mass / Molecular Mass.
Ammonium Chloride
0.1 = x / 53.49 (Found through periodic table)
.'. Mass = 5.35g
Sodium Hydroxide
0.1 = x / 40
.'. Mass =4g.
PROCEDURE
1. Obtain 5.35g of Ammonium Chloride and 4g of Sodium Hydroxide
2. Measure 100mL of water
3. Pour water into paper cup
4. Read the initial temperature of water
5. Place the ammonium chloride into water and stir until all dissolved using the thermometer as a stiring rod
6. Read final temperature
7. Repeat steps 4,5,6 using Sodium hydroxide
8. Calculate.
The main objective of the above is to find the change in temperature
OR ∆T.
∆T = ( Tf - Ti )
∆T = ( Final temperature - Initial Temperature)
Ammonium Chloride
∆T = 9 - 13
∆T = -4
Sodium Hydroxide
∆T = 21.25-9
∆T = 12.25
-----------------------------------
∆H=-mC∆T
We've got to find the pro numerals of the equation
For ammonium chloride
m = 0.1L
C = 4.18 (Specific heat capacity of water)
∆T = -4
.'. ∆H=-(0.1)(4.18)(-4 )
∆H= 1.672
Its positive .'. endothermic
i cbff now lol
same shit with sodium hydroxide.........
then you gotta find the molar heat of solution..
So taking Ammonium Chloride in example
∆H= 1.672
This is where i got lazy in the prac. This might actually take 80mins....it's soo long UGHH !! LFMAO! IM DEAD I NEED TO DO MY MATHS HOMEWORK !!
okokokok.
So remember we want 0.1n ?
This is the part where im not so comfortable in...
But heat of solution = j/mol^-1
sooooooo we got 0.1 moles...and the energy is -89.452
therefore
1.672
----------
0.1
.'. molar heat = 16.72
GOOODLUCKJJJJJJJJJJJJ
