Rozpuszczono 6,2 g tlenku sodu i otrzymano 400 cm3 roztowru. Oblicz stężenie molowe roztowru.
M(Na2O)=62g
m(Na2O)=6,2g
M(NaOH)=23g+16g+1g=40g
Vr=400cm³=0,4dm³
Na2O+H2O-->2NaOH
6,2g-------------xg
62g--------------2*40g
xg=6,2g*80g/62g
xg=8g
n=ms/M
n=8g / 40g/mol
n=0,2mol
Cm=n/Vr
Cm=0,2mol/0,4dm³
Cm=0,5mol/dm³
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M(Na2O)=62g
m(Na2O)=6,2g
M(NaOH)=23g+16g+1g=40g
Vr=400cm³=0,4dm³
Na2O+H2O-->2NaOH
6,2g-------------xg
62g--------------2*40g
xg=6,2g*80g/62g
xg=8g
n=ms/M
n=8g / 40g/mol
n=0,2mol
Cm=n/Vr
Cm=0,2mol/0,4dm³
Cm=0,5mol/dm³