AP Chemistry Problems





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Discus: SAT/ACT Tests and Test Preparation: May 2004 Archive: AP Chemistry Problems
By Starryqt22 (Starryqt22) on Monday, May 10, 2004 - 07:49 pm: Edit

These questions are from one of those AP practice books and I cannot figure out how to do them! If anybody can help, please do. Thanks in advance!

1) 3.00 mol of neon in a flask has a pressure of 1.50 atm. The pressure rises to 4.50 atm when 1.00 mol of hydrogen and some oxygen are added. How many moles of oxygen are added?
-The answer is 5.00, but what procedures do you go through to get these answers?
2) It takes 40.0 mL of 0.100 M NaOH to titrate 488 mg of a solid monoprotic acid to the phenolphthalein endpoint. What is the molecular mass of the acid?
-The answer is 122, once again how do you get this answer?
3) What is the molar mass of a nonelectrolyte that causes water to boil at 101.0 degrees Celsius when 23.3 g is dissolved in 100 g of distilled water (Kb for H2O = 0.51 degress Celsius/m)
-Answer is 116...once again how do you get this answer
Finally, 4) The boiling point of bromine is -7.2 degrees celsius, and its heat of vaporization is +15 kJ/mol. What is the entropy change when 1 mol of Br2(l) is converted to Br2(g) at -7.2 degrees celsius?
-Answer is 56.4 J/K

By Daromanian (Daromanian) on Monday, May 10, 2004 - 08:13 pm: Edit

1) 3 mol = 1.5 atm
new total = 4.5 atm.
that means there are 9 mol total in the flask.
9 mol - 3 mol N - 1 mol H2 = 5 mol.

2) ehh i hate acids.

3) diff in temp = molality * constant.
(23.3/molar mass)/.1L*.51 = 1
23.3/molar mass = .2
molarmass = 116

4)dG = 0 = 15000 - (-7.2+273)(dS)
i think

By Rono_G (Rono_G) on Sunday, May 16, 2004 - 01:37 am: Edit

aight, this is past the AP Chem testing deadline... but this is how you do #2...

Given:

Volume NaOH = 40.0 mL (0.04 L)
[ ] NaOH = 0.100 M
Mass Acid = 488 mg (0.488 g)

Find:

Molar Mass of Acid

To titrate something to its equivalence point, m1v1 must equal m2v2 - in other words, moles of acid must equal moles of base (and vice versa). Thereby, as we are given the m1 and v1 of the NaOH required to titrate the acid, we can calculate the moles of the NaOH -

moles = [ ] X V
= 0.04 X 0.1
= 0.004 moles

this MUST be the moles of acid present, thus

molar mass = grams/moles
= 0.488g/0.004 moles
= 122 g/mol

yeah,
thats pretty much it


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