Ile gramów wodorotlenku sodu znajduje się w 50g roztworu, o stężeniu 0,2 mol/dm³ i gęstości 1,275g/cm³?
V=m/d
m=50g
d=1,275g/cm^3
V=50g/1,275(g/cm^3)
V=39,21cm^3
n moli -> 39,2 cm^3
0,2 moli -> 1000cm^3
n = 0,00784 moli
M NaOH=40g/molm=n*Mm=0,00784*40m = 0,314 g NaOH
m=50gd=1,275g/cm3v=m/dv=50/1,275v = 39,2cm30,2mol -------------1000cm3xmol ------------------ 39,2cm3x = 39,2cm3 razy 0,2mol / 1000cm3 = 0,00784molaMNaOH=40g/moln=0,00784m=n razy Mm=0,00784 razy 40m=0,314g
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V=m/d
m=50g
d=1,275g/cm^3
V=50g/1,275(g/cm^3)
V=39,21cm^3
n moli -> 39,2 cm^3
0,2 moli -> 1000cm^3
n = 0,00784 moli
M NaOH=40g/mol
m=n*M
m=0,00784*40
m = 0,314 g NaOH
m=50g
d=1,275g/cm3
v=m/d
v=50/1,275
v = 39,2cm3
0,2mol -------------1000cm3
xmol ------------------ 39,2cm3
x = 39,2cm3 razy 0,2mol / 1000cm3 = 0,00784mola
MNaOH=40g/mol
n=0,00784
m=n razy M
m=0,00784 razy 40
m=0,314g