Molarity, Molality and Normality - Experiment No-2- Chemistry- CHEM 4202- Chemistry Lab

Subject: Chemistry Lab
Topic: Experiment no-2
Subject Code: CHEM  4202
Teacher Name: Md. Jahedul Islam

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Molarity ,  MOLALITY & NORMALITY

Solubility & Concentration

General Properties of Aqueous Solutions

• A solution is a homogeneous mixture of two or more substances.
• The substance present in the largest amount (moles) is referred to as the solvent.
• The other substances present are called the solutes.
•  A substance that dissolves in a particular solvent is said to be soluble in that solvent.

Solutions 

When the solvent is water the solution is said to be aqueous

Dilution
Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution.

 

Concentration of Solutions

Dilution is the process of preparing a less concentrated solution from a more concentrated one.

 

Another Dilution Problem

If 32 mL stock solution of 6.5 M H2SO4 is diluted to a volume of 500 mL. What would be the resulting concentration?

Solution: 

M1*V1   =  M2*V2

(6.5M) * (32 mL)  =  M2 * (500.0 mL)

M2  = (6.5 M * 32 mL)/500 mL 

M2  =    0.42 M

Concentration of Solution
Molarity  (M) = (Moles of solute)/(Liter of solution) = mol/ L





Molality (m) = Moles of solute / Kilograms of solvent
 

Molarity (M), or molar concentration, is defined as the number of moles of solute per liter of solution.

Calculating Molecular Weight

We can calculate the molecular weight of a substance using its chemical formula and the periodic table.

Say that we want to calculate the molecular weight of water. The molecular weight is the sum of the atomic weights of the atoms in the molecule.

Water contains hydrogen and oxygen. Then we write down how many of each of these atoms the molecule contains, which, in the case of water, is 2 and 1.

                                                                                                                 

For hydrogen, the atomic weight is 1.00794, and for oxygen it is 15.9994. Since there are 2 hydrogens in the molecule, the total weight of hydrogen in water is 2 times 1.00794, or 2.01588. There is only 1 oxygen, so the total weight of oxygen is 15.9994.

Finally, we add up the weights of all the atoms to get the total molecular weight of water, 18.0153 grams per mole.

Element	Atomic Weight (g/mol)
Ba	137.327
C	12.0107
Ca	40.078
Cr	51.9961
H	1.00794
Mg	24.3050
N	14.0067
Na	22.989770
O	15.9994
P	30.973761

Example:  12.6 g of NaCl are dissolved in water making 344mL of solution.  Calculate the molar concentration.

M= (moles solute/ L solution)

Example:  2    5.7 g KNO3 dissolves in a 233 mL solution.  What is molarity

1. Convert grams to moles

5.7 g KNO3   1 mol KNO3      = 0.056 mol KNO3

  101.10 g KNO3

2. Convert mL to L

233 mL  1 L       =  0.233 L

  1000 mL 

3. Divide mol/L

0.056 mol KNO3

0.233 L            

= 0.24 M KNO3

Molality (M) 
Molality (M), is defined as the number of moles of solute per kg of

Molality Examples

Example #1 - Suppose we had 1.00 mole of sucrose (it's about 342 grams) and proceeded to mix it into exactly 1.00 liter water. What would be the molality of this solution? 

Notice that one liter of water weighs 1000 grams (density of water = 1.00 g / mL and 1000 mL of water in a liter). 1000 g is 1.00 kg, so:

 

The answer is 1.00 mol/kg. 

Notice that both the units of mol and kg remain. And never forget this: replace the m with mol/kg when you do calculations. The m is just shorthand for mol/kg.

Example: 

Suppose you had 2.00 moles of solute dissolved into 1.00 L of solvent. What's the molality?

 

The answer is 2.00 m. 

Notice that no mention of a specific substance is mentioned at all. The molarity would be the same. It doesn't matter if it is sucrose, sodium chloride or any other substance. One mole of anything contains 6.02 X 1023 units.

Example - What is the molality when 0.75 mol is dissolved in 2.50 L of solvent?

The answer is 0.300 m.

Practice Problems

1)Calcuate the molality when 75.0 grams of MgCl2 is dissolved in 500.0 g of solvent.

2) 100.0 grams of sucrose (C12H22O11) is dissolved in 1.50 L of water. What is the molality?

3) 49.8 grams of KI is dissolved in 1.00 kg of solvent. What is the molality?

 

 

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