Here are few noteworthy points to keep in mind concerning this topic📋
EQUATION 1
Molarity (mol dm-3) = moles of solute (mol) ÷ volume of soltion (dm3)
1 dm3 = 1000 cm3
🧐Pay attention to the units❗️ ❕
EQUATION 2
Moles = Mass ÷ Molar Mass
Time for the examples❗️ ❕
Example 1🤓
5.78 g of CaCl2 are dissolved in water and the solution is made up to 250 cm3. Calculate the molarity of the solution.
(Relative atomic masses: Cl= 35.5, Ca= 40.1)
Step 1- Find the number of moles of CaCl2
Molar Mass of CaCl2 = (40.1+35.5+35.5) = 111.1 g mol-1
Moles of CaCl2= 5.78 g ÷ 111.1 g mol-1 = 0.0520 mol
Final Step- Find the molarity by substituting the correct values in EQUATION 1
Volume = 250 ÷ 1000 = 0.25
Molarity= 0.0520 ÷ 0.25 = 0.208 M
Example 2🤓
What is the mass of Na2SO4 required to prepare 1.5 dm3 of 1.6 mol dm-3 ?
(Relative atomic masses: O=16, Na= 32.1, S=32.1)
It can be seen that the question has provided the volume and molarity.
So the number of moles can be found
Step 1- Find the moles of Na2SO4 by REARRANGING EQUATION 1
Moles of Na2SO4 = Molarity × Volume = 1.5 × 1.6 = 2.4 mol
Last Step- Find the mass of Na2SO4 by REARRANGING EQUATION 2
Molar Mass of Na2SO4 = 142.1 g mol-1
Mass of Na2SO4 = Moles × Molar Mass = 2.4 × 142.1 = 341 g
Example 3🤓
Calculate the amount, in moles, of 25cm3 of hydrochloric acid (HCl) with a concentration of 2 mol dm3
Step 1- Convert to the correct unit
Volume = 25 ÷ 1000 = 0.025
Final Step- Find the moles of HCl by REARRANGING EQUATION 1
Moles of HCl = Molarity × Volume = 2 × 0.025 = 0.05
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