Ile gramów KNO3 (ms=) jest potrzebne do sporzadzenia 0,5dm3 roztworu o steżeniu molowym Cm=0,25mol/dm3
V = 0,5dm³
c = 0,25mol/dm³
n = cV
n = 0,5dm³ x 0,25mol/dm³ = 0,125mol
masa molowa
39 + 14 + 3 x 16 = 101 [g/mol]
101g/mol x 0,125mol = 12,625g
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V = 0,5dm³
c = 0,25mol/dm³
n = cV
n = 0,5dm³ x 0,25mol/dm³ = 0,125mol
masa molowa
39 + 14 + 3 x 16 = 101 [g/mol]
101g/mol x 0,125mol = 12,625g
----------------------------------------------------------------------------------------------------
Litterarum radices amarae sunt, fructus iucundiores
Pozdrawiam :)