Fill in the blanks in the table below (for each chemical, you need to show the det culations or derivations in your answer sheet; only presenting results in the table ceptable).
1) Na+
Atomic weight = 23 g/mol
Equivalent weight = 23 g/mol
n (eq/mol) = Mol wt/ Eq wt = 23/23 =1
Concentration= 23 mg/L
Concentration in mM = moles * 1000
= (Mass g/Mol Wt g/mol) * 1000
= (23 * 10-3 g/ 23 g/mol) * 1000
= 1 mM
Concentration in meq/L =Equivalents * 1000
= (W g/Eq Wt g/eq) * 1000
= (23 * 10-3 g/ 23 g/eq) * 1000
= 1
mg/L as CaCO3 = 23 mg/L * 2.18 = 50.14 mg/L as CaCO3 (2.18 is the conversion factor)
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