K1.922 x 10 x 3 mol K*5.77 x 10 mol/L 5.77 x 104 Mfolarity 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 millionppmppmM% =M. w x 1000N =.Eq. w x 1000+2solweight of will have to be in 1L of to a 100 ppm Pb soExample: (a) the molar conc. of 1 ppm each of Li and Pb. (b)ppm

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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

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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Concept explainers

Solutions

Solutions can be considered as homogeneous mixtures which contain two or more than two chemical substances. The chemical substance which is the greater part is defined as solvent. And the chemical substance that is dissolved in a solvent is referred to as solute. Solution chemistry plays an important role in chemistry since it supports many commercial applications of solutions.

Question

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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