P h i i l l a d e l l p h i i a U n i i v e r s s i i t t y
Lecturers : Dr. Ibrahim Khatib
And Dr.Zaher Algharaibeh
F a c u l l t t y o f f P h a r m a c y
D e p a r t t m e n t t o f f P h a r m a c e u t t i i c a l l S c i i e n c e s
Student Name: Student Number:
Course Name: Pharm Anal. Chem.
First Semester : of Academic Year 2014 / 2015
Course Number: 0510113 Sections : Date : 5/4 / 2015
First Exam Time : 50 minutes
Information for Candidates
1.
This examination paper contains 4 pages and 7 questions, totaling..20… mark.
2.
The marks for parts of questions are shown in round brackets.
Notes to Candidates
1. You should attempt all questions.
2. You should write your answers clearly .
Question 1 choose the correct answer (4 marks)
1.
The definition of the molarity is: a.
The molarity is the number of moles per kilogram solvent. b.
The molarity is the atomic weight expressed in gram. c.
The molarity is the solution that contain solution s one kilogram of solute. d.
The molarity the number of moles per liter
2.
A student was asked to make up a 10% w/w solution of sodium carbonate. Which one of the following methods should have been followed? a.
Dissolve 10 g of the sodium carbonate in water and make up to a mass of 100 grams by adding more water. b.
Dissolve 10 g of the sodium carbonate in water and make up to a volume of 1litre in a volumetric flask by adding more water. c.
Dissolve10 g of the sodium carbonate in water and make up to a volume of 100 cm 3 in a volumetric flask by adding more water. d.
Dissolve 10 g of the sodium carbonate in exactly 100 cm
3
of water
3.
If 10 mL of 0.1M solution HNO
3
was diluted to a volume of 50 mL what is the new concentration of the HNO
3 solution? a.
0.0002 M b.
2 M c.
0.002 M d.
0.02 M
4.
The system has gone from being ACID, past the equivalent point to the BASIC
(excess base), and back to the equivalent point again. The final titration to the equivalent point is called. a.
Direct titration b.
Acid base titration c.
back titration d.
a and b
1

Question2
Write four of the requirements for successful volumetric titration. (2 marks)
1 rapid titration
2.stoichiometric reaction
3.no side reaction
4.must have some indicators for end point
Question3
Calculate the mass of the following substances: a . 25 mL of 0.35 M Na
2
SO
3
solution ( Mw = 142.) (1 marks) mass = 0.025 x0.35 x 142 = 1.2425 gm b. 1 milliliter of 0.15 LiCl solution (Mw 42.5) (1 marks) mass = 0.0001x 0.15 x 42.5 =0.006375gm
c. 500ppm (Li (A.w7) (1 mark)
mass = 500 mg = 0.5 gram
Question4
Calculate the molar concentration of 20.00 ppm solutions K(Mw =39.1) and CH
4
(Mw 16) (3 marks)
0.2/39.1 = 5.11 x10 -4
0.2/16 = 1.25x10
-3
2

Question5
How many milliliters of concentrate HClO
4
(110.5) 70% (g/100 gram solution, density 1.35 g/cm3, are required to prepare 1 liter of 0.100 M solution? (3 marks)
Suppose we have liter solution of HClO
3
solution
Mass of liter solution =1000 x1.35 = 1350 gm
Mass of pure HClO
3
= 1350 x 0.70 = 945 gm
Number of moles in liter solution = 945/110.5 = 8.55
M = 8.55 /1 = 8.55M
M1V1 =M2V2
1000 X 0.1 =8.55 V2
V2 = 11.71 ml
Question 6
Distinguish between qualitative analysis and quantitative analysis? (2 points)
Qualitative analysis deal with identification of substances
Quantitative analysis deal with how much present in substance( concentration mass
–etc)
Question7
90.0 mL of 0.2105 M HCl acid was added in excess to 2.000 g calcium carbonate.
( Mw 100)The excess acid was back titrated with 0.1055 M sodium hydroxide. It required 12.2 mL of the base NaOH to reach the end point.
Calculate the percentage (w/w) of calcium carbonate in the sample.
(3 marks)
2HCl + CaCO
3

CaCl
2
+ CO
2
+ H
2
O ------ 1
HCl + NaOH

NaCl + H
2
O ------- 2
Mmoles of HCl = 90 x 0.2105 = 18.945
Excess = 12.2 x 0.10055 x1/1 = 1.2871
18.945 – 1.2871 = 17.658
Mmoles of CaCO
3
= 17.658 X 0.5 = 8.829
Percentage =8.829 x 100/2000 x100% = 44.14%
3