Concentration, exercise 4 (st-94) Different calculations round to The Concentration. To the 150 Milligrams One of them Lthe island with c = 0.3 mol/l will be 200 Milligrams Water given. What ? Concentration To the 300 Milligrams One of them Lthe island with c = 0.3 mol/l will be 200 Milligrams One of them (equal to) Lthe island with c= 0.5 mol/L given. What ? Concentration has The Lthe island Now .? Given be 1.5 L One of them Cooking salt solution with c = 0.15 mol/L. Well, you know what? to be But 2.0 L One of them Cooking salt solution with c = 0.3 mol/L Manufacture will be. How many Grams Fresh or chilled and Water must to the (c) Cooking salt solution Not yet Additionally Added will be.

  Aufgabe: st-94    

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Different calculations round to The Concentration.

1. To the 150 Milligrams One of them Lthe island with c = 0.3 mol/l will be 200 Milligrams Water given. What ? Concentration
2. To the 300 Milligrams One of them Lthe island with c = 0.3 mol/l will be 200 Milligrams One of them (equal to) Lthe island with c= 0.5 mol/L given. What ? Concentration has The Lthe island Now .?
3. Given be 1.5 L One of them Cooking salt solution with c = 0.15 mol/L.
   Well, you know what? to be But 2.0 L One of them Cooking salt solution with c = 0.3 mol/L Manufacture will be. How many Grams Fresh or chilled and Water must to the (c) Cooking salt solution Not yet Additionally Added will be.

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Lösung:

1. c = n/V Responsibility. n = c·V
       n(Lösung) = c·V = 0.3 mol/l·0.15 l = 0.045 mol
       new Volumes: 150 Milligrams + 200 Milligrams = 350 Milligrams Responsibility. 0.35 L
       c = n/V = 0.045 mol / 0.35 L = 0.129 mol/L
2. c = n/V Responsibility. n = c·V
       n₁(Lösung) = c₁·V₁ = 0.3 mol/l·0.3 l = 0.09 mol
       n₂(Lösung) = c₂·V₂ = 0.5 mol/l·0.2 l = 0.1 mol
       new n: n₁ + n₂ = 0.19 mol
       new Volumes: 300 Milligrams + 200 Milligrams = 500 Milligrams Responsibility. 0.55 L
       c = n/V = 0.19 mol / 0.5 L = 0.38 mol/L
3. Additional Volumes: 2.0 L – 1.5 L = 0.5 L
       n₁(1.5 L Lösung) = c₁·V₁ = 0.15 mol/L·1.5L = 0.225 mol
       n₂(2.0 L Lösung) = c₂·V₂ = 0.3 mol/L·2L = 0.6 mol
       with n=m/M Responsibility. m(NaCl) =n·M(NaCl)
       M(NaCl) = 58.5 g/mol
       m₁(NaCl) =0.225 mol · 58.5 g/mol = 13.16 g
       m₂(NaCl) =0.6 mol · 58.5 g/mol = 35.1 g
       Additional Quantity NaCl = 35.1-13.16 = about. 22 Grams

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