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Presentation On Stoichiometry: Mole And Solution Chemistry

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Published in: Chemistry | Science
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This PPT is suitable for the IGCSE and O Level Chemistry students. It only provides theoretical background of the topic. The practice questions, worksheets and topical questions will be discussed during the tutoring sessions

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  1. Moles and Solution Chemistry Chemistry (IGCSE / O-Level)
  2. What will you learn in this lesson? • Concentration of solutions and their calculations involving mole • Acid-base titration for finding unknown concentration of solution
  3. Concentration of Solutions When a chemical substance (the solute) is dissolved in a volume of solvent, we can measure the quantity of solute in two ways. • By mass (in grams) • By amount (in moles) The final volume of the solution is measured in cubic decimetre (dm3) • Mass concentration (g/dm3) • Molar concentration (mol/dm3) concentration = volurne «ihltion
  4. Concentration of Solutions The mass concentration of solution is measured in grams per cubic decimetre (g/dm3) The molar concentration of solution is measured in moles per cubic decimetre (mol/dm3) When a 1 mole of substance is dissolved in water and the solution is made up to 1 dm3 (1000 cm3), a solution with a concentration of 1 mol/dm3 is produced 1 mol of copper sulfate, CuSOa 2 mol of copper sulfate, CuSOa dissolve to make I dm• of solution, Concentration dissolve to make 2 cim• of solution. concentration dissolve to make I dm• Of solution, concentration — 2 mol/dm3 dissolve to make 2 dm• of solution. concentration making copper(ll) sulfate solutions of different concentraticms.
  5. Calculation using Solution Concentration number of moles in solution = molar concentration X volume of solution (in dm3) moles volume/ n centratioru' dm3 molar concentration number of moles in solution — x volume of solution (in cmg) 1000 moles ncentrabon volume/
  6. Calculation using Solution Concentration (contd.) Calculate the concentration of a solution of sodium hydroxide, NaOHr that contains 10 g of NaOH In a final volume of 250 cm3 Step 1. Find out how many moles of NaOH are present: relattve formula mass of naOH = 23 + 16 + 1 = 40 number of moles NaOH = Step 2. Ftnd the concentr*on: number of moles = = 0.25 mol x volume (in cm3) 0.25 = x 250 concentration = 0-25 = 1 mow—
  7. Calculation using Solution Concentration (contd.) NOW find the molar concentration of a solution of 14.3 g of hydrated sodium carbonate, NazC03TIOHzO in 500 crn3 of distilled water.
  8. Acid-Base Titration • Standard solution: A solution whose concentration is known precisely — this solution is then used to find the concentration of another solution using titration • Titration: A method of quantitative analysis using solutions: one solution is slowly added to a known volume of another solution using a burette until an end-point is reached. One of the solutions is acidic while the other is basic or alkali solution
  9. Acid-Base Titration (contd.) alkah scåut•cn accurat+. 25 cm J Of alkali a conical "ask - add indcator. Put acid in bwette — out - several timm. and take Run acid into flask colou —
  10. Acid-Base Titration (contd.) the ancetratton of a hydrochloric add solution A is against a standard sodium hydroxide is Found 20.0 cm3 •f add neanrat— 250 cm3 of O. 10 mob'dm3 NaOH solution. What Is the concentraton of the I. use Iomrmaeon the standard solution. How many motes of alkan ara tn the flagk? of uaOH x vak•me a-n3) = 2.5 x mol 2. Use he How many molgnof used? The * ua0H * H20 1 mol 1 mol 1 mal of neutratiys I mol OfHCI and so: 2.5 x 10-3 mol of Na0Hneutra11ses 2.5 x no-3 mol of 3. fie titr*on value. Wha is the concenu•ation of tf» acid? The acid contains 2.5 x 10-3 mol In 20.0 cm3. add 2.SxIO—3 x 1000 f 2W 0.125 mo.Vdm3
  11. Acid-Base Titration (contd.) Sulfuric acid can be neutralised using sodium hydroxide solution. H2S04(aq) 2Na0H(aq) — Na2S04(aq) + 25.0 cm3 of a sulfuric acid solution of concentration 0.2 mol/dm3 reacted with 10.0 cm3 of sodium hydroxide. Calculate: a the number Of moles Of Ifuric acid u sed b the number of moles of NaOH reacted c the concentration of the NaOH solution in mol/dm3.
  12. Balanced Equation acts as 'Footbridge' volume volume of gas solutron mass of product 5.22: A summary of the different ways in which a balanced equation acts as a 'footbridge' in Cal ons.
  13. Summary of Mole in Calculation concentration (mol/dmj x volume (drn3) I = IOOOcm3 amount (mol) number of particles 6.023 x M, (or A) lume + 24 dm3 (at r.t.p.)
  14. 7 8 9 Practice Questions Calculate the number of moles of gas there are in: a 480 cm3 of argon b 48 dm3 of carbon d ioxide c 1689 crn3 of oxygen. Calculate the volume (in cm3) of the following at r.t.p. a 1.5 moles of nitrogen b 0.06 moles of ammonia c 0.5 moles of chlorine. Calculate the concentration (in mol/dm3) of the following solutions. b d 1.0 mol Of sodium hydroxide is dissolved in distilled water to make 500 cm' of solution. 0.2 mol of sodium chloride is dissolved in distilled water to make 1000 cm3 of solution. 0.1 mol Of sodium nitrate is dissolved in distilled water to make 100 cm3 Of solution. 0.8 g of solid sodium hydroxide is dissolved in distilled water to a final volume Of I dm3. H = 1, 16, = 23)

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