In the earlier post, i mentioned that this topic Atoms, Molecules & Stoichiometry is usually taught revised in the 1st few weeks in Junior Colleges and Centralised Institute. In fact, the college teachers and lecturers do not really teach you but gives you questions to attempt – as a form of revision on a topic called Mole Calculations / Mole Concepts that you have learned in GCE O-Level Chemistry or other Basic Chemistry course (Click HERE if you need a quick revision).
One of the most common type of questions are related to Concentrations of Solutions.
Always remember that a Solution is made of of Solute (the minor component and usually a solid) and Solvent (the major component and usually a liquid, with water being most common).
I.e. Solution = Solute + Solvent
E.g. Salt Solution = NaCl (Solute) + Water (Solvent)
We can make the Salt Solution more diluted by either adding more water to it or removing solute from it.
The opposite must be true. To make the Salt Solution more concentrated, we can either remove the water or add more solute to it.
Example:
a) What is the Concentration of the salt solution when 117g of NaCl is added into 500 cm3 of water?
b) What is the corresponding Molar Concentration in mol/dm3?
Suggested Answer:
Using the formulae related to Concentration of Solutions that we discussed previously;
a) Concentration of Salt Solution = Mass of NaCl (g) / Volume of Solution (dm3)
= 117 g / 0.5 dm3
= 234 g/dm3
b) Mole of NaCl = Mass of NaCl (g) / Mr of NaCl
= 117 g / 58.5
= 2 mol
Molar Concentration = Mole of NaCl (mol) / Volume of Solution (dm3)
= 2 mol / 0.5 dm3
= 4.00 mol / dm3
Hope you are learning something useful on the above example. I would love to hear from you.
I always believed you learn something by doing it, and not reading it. As such, here is an quick check question for you to try it out.
Quick Check 1:
How many moles of sodium ions, Na+, are present in 250 cm3 of 0.4 mol/dm3 of Sodium Sulfate, Na2SO4?
PS: Try out the question and leave your answers in the comment section below.
