Molecular Weight: the sum of atomic weights of all elements
in a compound.
Molar Solution: 1 gram molecular weight of a substrate dissolved in 1
liter of solvent.
Molarity Using a Solid:
| |
1. Sodium chloride has a formula NaCl. Here the atomic weight
of sodium is 23 and that of chloride is 35, hence, molecular weight of NaCl is
23 + 35 = 58.
To make 1 molar solution of NaCl one should dissolve 58 grams of NaCl in 1 liter.
Alternatively, to make 1 mM solution from 1 M solution, dilute 1 ml of 1M to
1 liter (1 ml of 1M solution + 999 ml of water).
2. Sodium carbonate has a formula Na2CO3- Atomic weight
of Na is 23; C = 12 and O = 16.
Hence, molecular weight of sodium carbonate
will be:
| |
2 x 23 = 46
1 x 12 = 12
3 x 16 = 48 |
| Total |
= 106 |
Hence, to make 1M solution of Na2CO3 one should dissolve
106 g of Na2CO3 in 1 liter of water and to make 1 mM solution
dissolve 106 mg in 1 liter water.
|
Molarity Using a Liquid:
The molarity of a solution can be determined by the following formula:
| Molarity |
= |
Specific Gravity (g/mL) * Assay Value (in decimal units) * 1000mL/L
Molecular Weight (g/mole) |
For example:
PX1837A is a 70% iso-Propyl Alcohol Solution. The solution as an assay specification of 68.0-72.0%. There is a
specific gravity specification of 0.872-0.883 g/mL at 20/20°C. The chemical formula is
CH3CHOHCH3, which corresponds to a molecular weight of 60.10 g/mole. Lot 6070 has a specific
gravity of 0.879 and as assay value of 69.8%.
Therefore, the Molarity is:
| Molarity |
= |
0.879 (g.mL) * .698 * 1000mL/L
60.10 (g/mole) |
| Molarity |
= |
10.209 moles/L |
Normal Solution:
1 gram equivalent weight of a substance dissolved in 1 liter of solvent.
To determine the amounts to be used, one must determine the equivalent weight of the
compound. Equivalent weight is:
Molecular weight ÷ Valency
(valency is the combining capacity of elements)
For example:
| |
In sodium chloride, on Na combines with one Cl. hence, for NaCl
molecular weight = equivalent weight.
However, in case of Na2CO3 we need two Na to combine with
one CO3. Therefore, for Na2CO3 equivalent
weight is 53. (Molecular weight ÷ Valency = 106 ÷ 2). To make 1
normal solution of Na2CO3, dissolve 53 g in 1 liter.
Note:
You may need different amounts of material to make 1 Normal (1N) and 1 Molar (1M)
solutions.
Always use volumetric flask to make molar and normal solutions. Do
not dry these flasks in an oven. |
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