Oblicz, ile gram chlorku potasu KCL potrzeba do sporządzenia 0,4 dm^3 roztworu o stężeniu 0,2 mol/dm^3.
c = n/V
n= cV
n = 0,4dm³ x 0,2mol/dm³ = 0,08mol
m molowa KCl = 39g/mol + 35,5g/mol = 74,5g/mol
1 mol 74,5g
------=---------
0,08mola x
x = 5,96g
Pozdrawiam Marco12 :)
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c = n/V
n= cV
n = 0,4dm³ x 0,2mol/dm³ = 0,08mol
m molowa KCl = 39g/mol + 35,5g/mol = 74,5g/mol
1 mol 74,5g
------=---------
0,08mola x
x = 5,96g
Pozdrawiam Marco12 :)