c) Copper(II)sulfate is soluble in water. A student dissolved 25.0g of Copper(II)sulfate in 100mL of water in a polystyrene beaker stirring all the time. The temperature fell by 3 degree celsius. I. Calculate the enthalpy of solution of Copper(II)sulfate. (specific heat capacity of water = 4.18J/goC) Ii. suggest one source of error in this experiment and explain how the error affects the results.
Answers
Explanation:
Here's what I got.
Explanation:
The idea here is that you can use the heat absorbed by the solution to find the heat given off by the dissolution of the salt.
More specifically, you can assume that
Δ
H
diss
=
−
q
solution
The minus sign is used here because heat lost carries a negative sign.
To find the heat absorbed by the solution, you can use the equation
q
=
m
⋅
c
⋅
Δ
T
−−−−−−−−−−−−−
Here
q
is the heat gained by the water
m
is the mass of the water
c
is the specific heat of water
Δ
T
is the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample
As the problem suggests, you can approximate the mass and the specific heat of the solution to be equal to those of the pure water sample.
The temperature increases by
0.121
∘
C
, so you know that
Δ
T
=
0.121
∘
C
→
positive because the final temperature is higher than the initial temperature
Plug in your values to find
q
=
125
g
⋅
4.18 J
g
−
1
∘
C
−
1
⋅
0.121
∘
C
q
=
63.22 J
So, you know that the solution absorbed
63.22 J
, which implies that the dissolution of the salt gave off
63.22 J
. In other words, you have
Δ
H
diss
=
−
63.22 J
Convert the mass of sodium hydroxide to moles by using the compound's molar mass
2.4
⋅
10
−
4
g
⋅
1 mole NaOH
39.997
g
=
6.00
⋅
10
−
6
moles NaOH
You know that the enthalpy of dissolution when
6.00
⋅
10
−
6
moles of sodium hydroxide are dissolved in water, so use this info to find the enthalpy of dissolution when
1
mole of the salt dissolves
1
mole NaOH
⋅
−
63.22 J
6.00
⋅
10
−
6
moles NaOH
=
−
1.054
⋅
10
7
J
Finally, convert this to kilojoules
1.054
⋅
10
7
J
⋅
1 kJ
10
3
J
=
1.054
⋅
10
4
kJ
Therefore, you can say that the enthalpy of dissolution, or molar enthalpy of dissolution, for sodium hydroxide is
Δ
H
diss
=
−
1.1
⋅
10
4
.
kJ mol
−
1
−−−−−−−−−−−−−−−−−−−−−−−−−−−
The answer is rounded to two sig figs, the number fo sig figs you have for the mass of sodium hydroxide.
SIDE NOTE The accepted value for the enthalpy of dissolution of sodium hydroxide in water at
25
∘
C
is
Δ
H
diss
=
−
44.51 kJ