Key Concepts
- weight/volume percentage concentration is usually abbreviated as w/v (%)
- w/v (%) = mass solute ÷ volume solution × 100%
- common units are g/100mL (%)
Example
What is the weight/volume percentage concentration of 200mL of aqueous sodium chloride solution containing 5g NaCl?
Calculate the weight/volume (%) = mass solute ÷ volume of solution × 100%
mass solute (NaCl) = 5g
volume of solution = 200mL
w/v (%) = 5g ÷ 200mL × 100%
= 2.5g/100mL
= 2.5%
Key Concepts
- volume/volume percentage concentration is usually abbreviated as v/v (%)
- v/v % = [(volume of solute)/(volume of solution)] × 100%
- volume percent is relative to volume of solution, not volume of solvent.
- liqud and gas volumes are not necessarily additive. If you mix 12 ml of ethanol and 100 ml of wine, you will get less than 112 ml of solution.
Example
Wine is about 12% v/v ethanol. This means there are 12 ml ethanol for every 100 ml of wine.
70% v/v ethnol can be prepared by taking 700 ml of ethonol and adding sufficient water to obtain 1000 ml of solution.
Key Concepts
- molar concentration is usually abbreviated as mol/L or M
- The concentration of a solution is usually given in moles per litre. This is also known as molarity.
- molar concetration = moles of solute ÷ volume of solution in litres (L)
Example
What is the molarity of a solution in which 3.57g of sodium chloride, NaCl, is dissolved in enough water to make 25.0ml of solution?
First, convert from mass of NaCl to moles of NaCl.
3.57g NaCl ÷ (58.44g NaCl/molNaCl) = 0.0611mol NaCl
25.0 mL = 0.025 L
molarity = 0.0611mol ÷ 0.025L = 2.44M NaCl
Key Concepts
- Dilution refers to make a lower concentration solution from higher concentrations.
- The dilution fomula is:
Concentration(stock) × Volume(stock) = Concentration(dilute) × Volume(dilute) - This equation is commonly abbreviated as: C1V1 = C2V2
Example
How many millilieters of 6M NaOH are needed to prepare 300 mL of 1.2 M NaOH?
6M × V1 = 1.2M × 300ml
V1 = 1.2M × 300ml ÷ 6M
V1 = 60mL
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