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solution of KCI (its v Normality. culate the weight of solu (a) 43.5 mL of O.175 N (b) 10 mL of 0.03 N KIL (c) 5 litres of 0.25 N K2

solution of KCI (its v Normality. culate the weight of solu (a) 43.5 mL of O.175 N (b) 10 mL of 0.03 N KIL (c) 5 litres of 0.25 N K2

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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K 1.922 x 10 x 3 mol K* 5.77 x 10 mol/L 5.77 x 104 M folarity and Normality. 2- Calculate the weight of solute in grams for each of the followings solutions: (a) 43.5 mL of O.175 N Hg(N03)2 (b) 10 mL of 0.03 N KI if the reaction product is Agl (c) 5 litres of 0.25 N K,CO, in acidic medium. The relationship between molarity, normality and part per million ppm ppm M% = M. w x 1000 N =. Eq. w x 1000 +2 sol weight of will have to be in 1L of to a 100 ppm Pb so Example: (a) the molar conc. of 1 ppm each of Li and Pb. (b) ppm
Transcribed Image Text:K 1.922 x 10 x 3 mol K* 5.77 x 10 mol/L 5.77 x 104 M folarity and Normality. 2- Calculate the weight of solute in grams for each of the followings solutions: (a) 43.5 mL of O.175 N Hg(N03)2 (b) 10 mL of 0.03 N KI if the reaction product is Agl (c) 5 litres of 0.25 N K,CO, in acidic medium. The relationship between molarity, normality and part per million ppm ppm M% = M. w x 1000 N =. Eq. w x 1000 +2 sol weight of will have to be in 1L of to a 100 ppm Pb so Example: (a) the molar conc. of 1 ppm each of Li and Pb. (b) ppm
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