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Lab #3

Lab #3. Solution and Dilution. Outline. -Concentration units Molar Concentration. Normal Concentration . - Dilution. 2-Molar Concentration. the mass required to prepare a solutions g = number of moles mol × molar mass mol/g Number of moles mol = concentration mol/L × volume L.

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Lab #3

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  1. Lab #3 Solution and Dilution

  2. Outline -Concentration units • Molar Concentration. • Normal Concentration. - Dilution

  3. 2-Molar Concentration • the mass required to prepare a solutions g = number of moles mol× molar mass mol/g • Number of moles mol= concentration mol/L × volume L

  4. Molar Concentration • Example: Calculate the mass of barium nitrate Ba(NO3)2 required to produce 100.0 mL of 0.100 mol/L solution. • List the given information: v = 100.0 mL C = 0.100 mol/L

  5. Identify the wanted information: Calculate m = of Ba(NO3)2 • Plan and perform the calculations: There is no formula that allows you to convert volume and molar concentration to mass however, volume and concentration can be converted to moles and moles in turn can be converted to mass using molar mass.

  6. The volume must be converted from mL to L. • Convert volume from mL to L:

  7. Calculate the number of moles of solute: • Calculate the molar mass of the solute: Ba(NO3)2

  8. Calculate the mass of the solute to be dissolved:  • Communicate the required mass: The mass of barium nitrate required to prepare 100.0 mL of 0.100 mol/L solution is 2.61 g.

  9. 3-Normal Concentration or Normality (N): • A normal is one gram equivalent of a solute per liter of solution. • Example: a 1 N solution of hydrochloric acid HCl would also be 1 N for H+ or Cl-ions for acid-base reactions.

  10. Equation of Normality • Weight of compound = volume (L) X normal concentration X (atomic weight / equivalent weight). • Volume = weight / density. Weight = wt. Volume = vol. Concentration = conc.

  11. Normality • Example: Prepare 1 L of 0.1 equilibrium Hcl, density= 11.68 g/ml, atomic wt =36.5, equivalent wt= 1. • Wt of compound= 1 X 0.1 X (36.5 / 1) = 3.65 g. • Vol. of Hcl = 3.65 / 11.68 = 0.3125 ml. • Vol. of solvent = 1000 - 0.3125 = 999.6875 • ml.

  12. Dilution • If lower concentrations than the one available are desired these can be made by dilution. A suitable amount of solution is pipetted into a clean flask and solvent added to the mark.

  13. Example: Prepare a 100 ml Hcl with 20% conc., density 1.14g/ml; by using other solution of 98% Hcl, density 1.84 g/ml. • Vol. X conc. (before dilution) = vol. X conc. (after dilution). • Weight = density X volume. • Wt of 20% Hcl =1.14 X 0.20= 0. 228 g. • Wt of 98% Hcl = 1.84 X 0.98 = 1.8032 g. • Vol. of 98% Hcl= vol. X con. (of 20% Hcl)/conc. of 98% Hcl. • Vol. of Hcl = 100 X 0. 228 / 1.8032 =12.6 ml. • Vol. of solvent = 100 – 12.6 = 87.4 ml.

  14. Example 2:Prepare a 100 ml Nacl with 1% conc., density 1.008g/ml; by using other solution of 10% Nacl, density 1.146 g/ml.

  15. Wt of 1% Nacl= 1.008*0.01=0.01008g. • Wt of 10% Nacl= 1.146*0.10=0.1146g. • Vol of 10% Nacl= (100* 0.01008)/ 0.1146 = 8.7958 mL • Vol. of solvent = 100 – 8.7958 = 91.2042=91 ml.

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