CHEM 103: FINAL EXAM STUDY GUIDE WITH ANSWERS

CHEM 103: FINAL EXAM STUDY GUIDE WITH
ANSWERS


 Water (H2O) with a molecular weight of 18 has a boiling point of 100oC
and methane (CH4) with a somewhat similar molecular weight of 16

has a boiling point of -161oC. This difference is much higher than can
be explained by the polarity of water. Explain what causes this
difference.
Hydrogen bonds which are strong intermolecular attractive forces
form between water molecules requiring much more energy to
separate the molecules than for other molecules.



 Show the calculation of the molar mass (molecular weight) of a
solute if a solution of 14.6 grams of the solute in 200 grams of water

has a freezing point of -1.35oC. Kf for water is 1.86 and the

freezing point of pure water is 0oC. Calculate your answer to 0.1
g/mole. Δtf = Kf ×m




∆tf = Kf x m




molality = ∆tf / Kf = 1..86 = 0.726 m




molality = (gsolute / MW) / (gsolvent / 1000)




0.726 = (moles) / ()

, CHEM 103: FINAL EXAM STUDY GUIDE WITH
ANSWERS


Moles = 0.726 x 0.200 = 0.1452




0.1452 = (14.6 / MW)




MW = 14..1452 = 100.6


Question 1

1. Convert 0.00346 to exponential form and explain your answer.


2. Convert 2.76 x 104 to ordinary form and explain your
answer. Your Answer:
1. 0.00346 = smaller than 1 = negative exponent, move decimal 3

places = 3.46 x 10-3




2. 2.76 x 104 = positive exponent = larger than 1, move decimal
4 places = 27600


Question 2

Do the conversions shown below, showing all work:

1. 28oC = ?
oK 2.158oF = ?

oC

3. 343oK = ? oF