Description
Please hand-write your results to show the correct procedure .Results without showing the procedure will not be considered in any case. (Always use correct significant figures)
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Explanation & Answer
Most of the solutions are typed and included in the word document. 4 of the problems are handwritten and attached individually. Please note that a few of the problems asked you to look up values in your text. I found those values online. They should not differ too greatly from what is in your text, but you may want to double check them. The values will not change the process, but would change the calculations just slightly.Please let me know if you have any questions about these problems.
CHEM-112-01 General Chemistry II lecture , MOCK FINAL
EXAM_1_ is for you to practice at home before FINAL , (11 +13
+14+15+16+17+19+ 20) Please hand-write your results to show
the correct procedure .Results without showing the procedure will
not be considered in any case. (Always use correct significant
figures)
1.
Which member of each of the following pairs of substances would you expect to
have a higher boiling point? SO2 and CO2, EXPLAIN your ANSWER
The boiling point of a compound will increase as the strength of the intermolecular forces
increases. SO2 is a polar molecule due to the 2 sets of unpaired electrons. That means
the forces will include dipole-dipole forces. CO2 is a non-polar molecule due to the
linear shape. A non polar molecule like CO2 has only London dispersion forces as
intermolecular forces. SO2 has the stronger intermolecular forces, so it will also have the
higher boiling point.
2.
List the types of intermolecular forces that exist between molecules (or basic
units) in each of the following species: benzene (C6H6) EXPLAIN your ANSWER.
There are 3 possible intermolecular forces: dipole-dipole forces between polar molecules,
London dispersion forces between all molecules and hydrogen bonding between polar
molecules where H is bonded to N, O or F.
Benzene is a non-polar molecule. There is not N, O or F to bond to the H in the benzene.
This means that the only intermolecular forces in benzene are London dispersion forces.
3.
A solution is prepared by condensing 4.00 L of a gas, measured at 27°C and 750.
mmHg pressure, into 59.0 g of benzene. Calculate the freezing point of this solution.
To calculate freezing point depression, you need to know the molality of the solution
(in moles of solute / kg solvent) and the depression constant for the solvent.
In this case the solvent is benzene which has kf value of 5.12 degC / m.
Since benzene is the solvent, its mass should be in kg instead of g.
59.0 g benzene = 0.059 kg benzene
Then we need to figure out the number of moles of the gas that is condensed into the
benzene. For this you can use the ideal gas law.
V = 4.00 L, P = 750 mmHg and T = 27 deg C + 273.15 = 300.15 K
I looked up the value of R with the correct units. If you only have one value of R
available you would need to convert the pressure and temperature units to the
appropriate one.
R = 62.36367 L mmHg / K mol
PV = nRT
PV
n=
RT
4.00 L∗ 750 mmHg
n=
L mmHg
62.36367
∗ 300.15 K
mol K
n= 0.1602696 mol gas
Now that the number of moles of gas has been determined you can find the molality
of the solution.
0.1602969 mol
0.059 kg
m= 2.71643445635 m
m=
Finally you can calculate the freezing point depress...