A7LAM CSCA ACADEMYCSCA Academy / Chemistry / Basic concepts and calculations
CSCA Chemistry · Lesson 3 of 21Solutions and concentration
Solutions et concentration · المحاليل والتركيز
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Most CSCA calculations happen in solution. Molar concentration tells you how many moles are in each litre.
Key ideas
- Molar concentration c = n/V (mol/L).
- Dilution keeps the number of moles: c₁V₁ = c₂V₂.
- Mass fraction w = m(solute)/m(solution) × 100%.
- To prepare a solution: weigh, dissolve, transfer to a volumetric flask, fill to the mark.
Formulas to know
- c = n / V
- c₁V₁ = c₂V₂
- n = m / M
Worked example
4.0 g of NaOH (M = 40 g/mol) is dissolved to make 500 mL of solution. Concentration?
- Moles: n = 4.0 / 40 = 0.10 mol
- Volume: V = 500 mL = 0.500 L
- Divide: c = 0.10 / 0.500 = 0.20 mol/L
Answer: 0.20 mol/L
Practice questions
1. 100 mL of 2 mol/L HCl is diluted to 500 mL. New concentration?
0.4 mol/L ✓ · 10 mol/L · 1 mol/L · 0.2 mol/L
2 × 100 / 500 = 0.4.
2. Moles of NaCl in 250 mL of 0.4 mol/L solution?
0.1 mol ✓ · 1.6 mol · 0.4 mol · 100 mol
0.4 × 0.250 = 0.1 mol.
3. 20 g of sugar dissolved in 80 g of water. Mass fraction?
25% · 20% ✓ · 80% · 16%
20 / (20 + 80) = 20%.