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Marizza181 [45]
3 years ago
7

Analyze the data and determine the actual concentration of calcium chloride in the solution. Show all calculations and report in

% wt/v concentration.
Known; Mass of CaCl2 present in original solution, based on actual yield= 1.77g moles
CaCl2 present in original solution, based on actual yield= 1.77g/molar mass of CaCl2=1.77g/110.98g/mol=0.016 moles
Total Volume of solution =V, which is 80ml
Chemistry
1 answer:
Rus_ich [418]3 years ago
6 0

Answer:

2.21% wt/v

Explanation:

The mass/volume percentage, %wt/v, is an unit of concentration used in chemistry defined as 100 times the ratio of the mass of solute in g (In this case, CaCl2 = 1.77g) and the volume of solution in mL = 80mL

The %wt/v of this solution is:

%wt /v = 1.77g / 80mL * 100

%wt/v = 2.21% wt/v

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. What mass of ammonium chloride must be added to 250. mL of water to give a solution with pH
Eduardwww [97]

The mass of ammonium chloride that must be added is : ( A ) 4.7 g

<u>Given data :</u>

Volume of water ( V )  = 250 mL = 0.25 L

pH of solution = 4.85

Kb = 1.8 * 10⁻⁵

Kw = 10⁻¹⁴

Given that the dissolution of NH₄Cl gives NH₄⁺⁺ and Cl⁻ ions the equation is written as :

NH₄CI  +  H₂O  ⇄  NH₃ + H₃O⁺

where conc of H₃O⁺

[ H₃O⁺ ] = \sqrt{Ka.C}   and Ka = Kw / Kb

∴ Ka = 5.56 * 10⁻¹⁰

Next step : Determine the concentration of H₃O⁺  in the solution

pH = - log [ H₃O⁺ ] = 4.85

∴ [ H₃O⁺ ] in the solution = 1.14125 * 10⁻⁵

Next step : Determine the concentration of NH₄CI in the solution

C = [ H₃O⁺ ]² / Ka

  = ( 1.14125 * 10⁻⁵ )² /  5.56 * 10⁻¹⁰

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Determine the number of moles of NH₄CI in the solution

n = C . V

  = 0.359 mol / L  * 0.25 L =  0.08979 mole

Final step : determine the mass of ammonium chloride that must be added to 250 mL

mass = n * molar mass

         = 0.08979 * 53.5 g/mol

         = 4.80 g  ≈ 4.7 grams

Therefore we can conclude that the mass of ammonium chloride that must be added is 4.7 g

Learn more about ammonium chloride : brainly.com/question/13050932

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