Solution :- Reactant-1 Potassium carbonate - Concentration (c) = 0·200 M = 0.200 mol/L Volume (V) = 20 mL = 20 - L = 0·02 L [1L = 1000 mL] 1000 Now, Moles(n) Concentration (2) = Volume (V) => Moles (n) = Concentration (C) x Volume (v) = 0.200 mol x 0.02 L = 0·004 mol. Molar mass = 147.01 g/mol. Now. Mass = Molex Molar mass = 0.004 molx 147.01 gm = 0.588 gm. Hence, 0.588 gm of potassium carbonate is used. Reactant-2- Calcium chloride dihydrate Concentration (e) = 0·220 M = 0·220 mol/L Volume (V) = 20 mL = 0.02 L Now, C = 꼼 ⇒ n = cx V = 0·220 mal x 0·02 L = 0.0044 mol. Molar mass = 138.21 gm/mol. Now, Mass= Molex Molay mass = 0.0044 mol x 138.21 gm -=0.608 gm. mol Hence, 0.608 gm of calcium chloride dishydrate is used.
Solution :- Reactant-1 Potassium carbonate - Concentration (c) = 0·200 M = 0.200 mol/L Volume (V) = 20 mL = 20 - L = 0·02 L [1L = 1000 mL] 1000 Now, Moles(n) Concentration (2) = Volume (V) => Moles (n) = Concentration (C) x Volume (v) = 0.200 mol x 0.02 L = 0·004 mol. Molar mass = 147.01 g/mol. Now. Mass = Molex Molar mass = 0.004 molx 147.01 gm = 0.588 gm. Hence, 0.588 gm of potassium carbonate is used. Reactant-2- Calcium chloride dihydrate Concentration (e) = 0·220 M = 0·220 mol/L Volume (V) = 20 mL = 0.02 L Now, C = 꼼 ⇒ n = cx V = 0·220 mal x 0·02 L = 0.0044 mol. Molar mass = 138.21 gm/mol. Now, Mass= Molex Molay mass = 0.0044 mol x 138.21 gm -=0.608 gm. mol Hence, 0.608 gm of calcium chloride dishydrate is used.
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter10: Solutions
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Problem 5QAP: Silver ions can be found in some of the city water piped into homes. The average concentration of...
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