Sch 101: Fundamentals Of Chemistry I Question Paper
Sch 101: Fundamentals Of Chemistry I
Course:Bachelor Of Science In Disaster Preparedness And Environmental Technology
Institution: Masinde Muliro University Of Science And Technology question papers
Exam Year:2012
Masinde Muliro University of Science and Technology (MMUST)
Course Title: Fundamental of Chemistry II
First year second semester
Course code: SCH 101
Time: 2 HOURS
Instruction to candidates:
Answer all questions
Question one:
a) Define the following Gas laws
i) Boyle’s Law
ii) Avogadro’s Law
iii) Dalton’s Law of partial pressures (3marks)
b) From the standpoint of the Kinetic molecular theory, discuss briefly the properties of gas molecules that cause deviation from ideal behavior (4marks)
c) At atmospheric pressure of a compound boils at 1200C. At 0.5 atmosphere pressure, will it boil at a higher or lower temperature? Explain (2marks)
d) A balloon has 20 grams of O2, 40 grams of CH4 and 10 grams of He at a Ptot of 600mmHg. What is the partial pressure of each gas? (3marks)
e) A sample of 2.05g of the plastic polystyrene was dissolved in enough toluene to form 100mL of solution. The osmotic pressure of this solution was found in 0.01194 atm at 250C. Calculate the molar mass of polystyrene. (R=0.08206L.atmK-1mol-1)
Question two
You are given a sketch of an electrochemical cell. Study the sketch and answer the following questions.
a) What type of cell is depicted in the sketch? (1mark)
b) What is the oxidizing agent in a cell? (1mark)
c) Write the equation for the overall reaction that occurs in the cell. (1mark)
d) Identify the anode in the cell. Justify your answer. (2marks)
e) Redraw the sketch; use an arrow to clearly indicate the direction of current flow, the key parts of the cell and their functions. (5marks)
f) Calculate the value of the cell potential and comment on the spontaneity of the redox reaction. (2marks)
g) From fundamentals of chemistry, it is given that ^G 0= -nFE, show that
E=E 0 – RT/nF In Q
(3marks)
( Fe 2+ (aq) + 2e- > Fe (s) E0 = -0.41V )
( Cd 2+ (aq) + 2e- > Cd (s) E0 = -0.40V )
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