In Mole Calculations, many students lose marks in questions related to Volumetric Analysis, especially on those that requires them to understand the concepts of CONCENTRATIONS of solutions. Volumetric Analysis are always a highlight in our annual Mole Calculations Mastery Workshop.
Concentrations of a solution refers to the amount of solute dissolved in 1 dm3 of the solution.
Do note that:
Solution = Solute (minor component) + Solvent (major component)
Usually in basic chemistry such as O Levels, the solvent used is Water. As such, the solution formed is usually aqueous solutions whereby certain solid or liquid are dissolved in water.
Concentrations can be expressed in:
- Grams of solute per dm3 (unit will be g/dm3)
- Moles of solute per dm3 (unit will be mol/dm3)
The most important formulae student need to know in volumetric analysis are:
Conc. (mol/dm3) = Amt. of Solute (mol) / Vol. of Solution (dm3)
Conc. (g/dm3) = Amt. of Solute (g) / Vol. of Solution (dm3)
Let’s check out a question to see how we can use the formulae above.
60g of NaOH is dissolved in water and make up to the total volume of 500cm3. Calculate the concentration of solution in i) g/dm3 and ii) mol/dm3.
Mass of NaOH = 60g
Volume of solution in dm3 = 500/1000 = 0.5dm3
Concentration of solution in g/dm3 = 60 g /0.5 dm3 = 120 g/dm3
Mr, Relative molecular mass of NaOH = 23 + 16 + 1 = 40
Moles of NaOH = Mass / Mr = 60/40 = 1.5 mol
Concentration of solution in mol/dm3 = 1.5 mol / 0.5 dm3 = 3 mol/dm3
Isn’t it easy if you know the strategy of doing it? I would love to hear from you. Leave me a comment.
For those itching to try out a question on your own. Check out the Quick Check below with a little bit of twist.
Quick Check 1:
Given concentration of a NaOH solution is 1.5 mol/dm3. How many grams of NaOH solute are contained in 2.0 dm3 of solution?
PS: Try it out and leave your suggested answer (working even better) below! Love to hear from you.