Does anyone know how to calculate normality of the following aqeous solutions?
.25 M Na C2H3O2 (sodium acetate)
.50 M K2SO4 (potassium sulfate)
5.0% (w/v) CaCl2 (calcium chloride)
please help??? i need this one ! please
2007-09-24 18:34:06 · 3 answers · asked by SJ 1 in Science & Mathematics ➔ Chemistry
someone please help me with this ??
2007-09-24 19:00:25 · update #1
Normality=Molarity when dissociation is complete....which is true for salts like these in solution. There is no way in hell they forgot to tel you this in the book...you need to read more carefully...
2007-09-24 18:39:08 · answer #1 · answered by Anonymous · 0⤊ 0⤋
For the first two, Normality= molarity
For CaCl2, normality = 2*molarity,
2007-09-25 01:40:33 · answer #2 · answered by cattbarf 7 · 0⤊ 0⤋
Na C2H3O2 MW= 82 g/L (g/M)
N=1
Nsolution= (M*N)= 0.25
K2SO4 MW= 176 g/L
N=2
Nsolution=1.0
CaCl2 - contains 5g of CaCl for every 100ml of solution.
MW=111 g/L
5g/100ml=x g/1000ml
x=50 g
M=111/50=2.22M
N=2
Nsolution=4.44
2007-09-25 03:24:35 · answer #3 · answered by e_coli_lover 1 · 0⤊ 0⤋