Sunday 14 January 2018

Chemistry MDCAT UHS-1A,Lec#4

Topics

1. Percentage composition
2. Molarity 
3. Mole fraction

Phase

Every sample of matter with uniform properties and a fixed composition is called a phase
 Examples 
i)   Sample of pure water is a phase 
ii)  Homogeneous mixture of glucose in water is a phase.

 SOLUTION 

A mixture of two or more kinds of different molecular or ionic substances is called solution.

 Components of a solution

 Solute 
i) In the case of solution a gas or solid dissolved in a liquid, the gas or solid is the solute.
ii) Generally, the component of a solution, which is in smaller amount, is called solute.
 Solvent
i) In the case of solution of a gas or solid in a liquid,the liquid is the sovent.
ii) Generally the component of solution which is in larger amount is called solvent.
Example 
When sodium chloride is dissolved in water, sodium chloride is the solute and water is the solvent.

CONCENTRATION UNITS OF SOLUTIONS

1. PERCENTAGE COMPOSITION

Solute in percent of solvent.
Four ways of expressions of percentage composition.
i - w/w %
it is weight of solute dissolved per 100 grams of solution.
ii - v/w %
 it is volume of solute dissolved per 100 gram of solution.
iii - w/v %
 it is a weight of solute in grams dissolved per 100 cm3 of solution.
iv - v/v %
 it is volume of solute in cm3 dissolved per 100 cm3 of solution.
 All above expressed by percentage (%)

2.Molarity

The number of moles of solute dissolved per dm3(1000cm3) or 1 litre of a solution is called molarity.
 symbol
 it is denoted by 'M'.
Examples
 0.1 M NaOH solution and 0.02 M KMnO4 solution.
 formulae
i) Molarity(M)= Moles of solute/volume of solution in dm3 = n/v
ii) Molarity(M)= Mass of solute/ molar mass of solute × 1/volume of solution in dm3 = W2/M2 × 1000/V(cm3)

3.MOLE FRACTION

(Especially suitable for solution having more than two components)
The ratio of the number of moles of a component to the total number of moles of all the components of solution is called mole fraction.
Symbol
it is denoted by 'X'.
Explanation 
Let us having components A, B and C making a solution and their number of moles are nA,nB and nC respectively. XA ,XB and XC are their mole fractions and are given below:-
         XA = nA / nA + nB + nC
         XB = nB / nA + nB + nC
         XC = nC / nA + nB + nC
 •  The sum of the mole fractions of all the components of solution is unity (one)
•   To get mole percent, mole fraction is multiplied by 100.
•   There are no formal units of mole fraction because it is a ratio of same quantity.
•   Mole fraction of a component in a solution is always less than 1.

A comparison of various concentration units

• Molarity is temperature dependent and represent in its units as moles/ dm3 
• Molality is temperature independent and represent in its units as moles/kg
• Mole fraction is temperature independent and has no units.
• Parts per million(PPM) is temperature independent and represent in its units as PPM.

Multiple Choice Questions (MCQs)

1. 5.0 gram NaOH is dissolved in 1000 cm3 water. The molarity of NaOH is
A. 0.125 M
B. 0.250 M
C. 0.500 M
D. 1.00 M
2. 250 cm3 of 0.2 molar Potassium Sulphate solution is mixed with 250 cm3 of 0.2 molar potassium chloride solution.The molar concentration of potassium ions is
A. 0.2 Molar
B. 0.3 Molar
C. 0.25 Molar
D. 0.35 Molar
3. The number of molecules of glucose in 500 cm3 of its 1 molar solution are
A. 1.8 × 10^23
B. 1.2 × 10^24
C. 3.1 × 10^23
D. 6.02 × 10^23
4. 9 gram glucose is dissolved in 45 gram water. The mole fraction of glucose in solution is
A. 1/5
B. 5.1
C. 1/55.5
D. 1/51
5. A solution has 46 gram ethanol, 48 gram methanol and 45 gram water. Mole fraction of ethanol in the solution is
A. 0.1
B. 0.2
C. 0.4
D. 0.5
 6. A solution of glucose is 5%.The volume in which 1g mole of it will be dissolved is 
A. 1.0 dm3
B. 1.6 dm3
C. 2.5 dm3
D. 3.2 dm3
7. When a liquid solute is dissolved in a liquid solvent then the best unit of concentration is
A. w/w %
B. w/v %
C. v/w % 
D. v/v %
8. How many grams of NaOH are present in 250 cm3 of its 0.2 M solution
A. 1.0 g
B. 2.0 g
C. 20 g
D. 40 g
9. When we dissolve 15.8 gram of KMnO4 in 500 cm3 of water. The solution is
A. 0.1 M
B. 0.2 M
C. 0.3 M
D. 1.0 M
10. Which one is temperature dependent
A. Molarity
B. w/ w %
C. Mole fraction
D. Molality

Answers

1. A
2. B
3. C
4. D
5. B
6. D
7. D
8. B
9. B
10. A

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