Answer:
0.0400M of KI
Explanation:
Molarity is an unit of concentration defined as the ratio between moles of solute and liters of solution.
When you add 10.0 mL of 0.10M KI and 15.0mL, total volume is:
25.0mL = <em>0.025L of solution</em>
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And moles of KI are:
0.0100L × 0.10M = <em>0.00100 moles of KI</em>
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Thus, molarity is:
0.00100 moles / 0.025L = <em>0.0400M of KI</em>
The given compound is Aluminum sulfate, Al2(SO4)3:
Molar masses:
Aluminum = 27 g/mol
Sulfur = 32 g/mol
Oxygen = 16 g/mol
The total molar mass is 342 g/mol
The ratio by mass of the elements:
Aluminum = 27*2/342
= 0.16
Sulfur = (32*3)/342
= 0.28
Oxygen = (16*12)/342
= 0.56
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0.0015 kilometers is for sure the answer!
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⁺ ] =
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
= 0.359 mol / L
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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