Oblicz stęzenie molowe roztworu jeżeli w 500 cm szesiennych roztworu znajduje się 8g NaOH.
Oblicz masę wody niezbędną do przygotowania 200g 7% roztworu.
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Zad.1.
Cm = ms/M*Vr
ms = 8g
M(NaOH) = 40 g/mol
Vr = 500cm³ = 0,5dm³
Cm = 8/40*0,5 = 0,4 mol/dm³
Zad.2.
Cp = ms*100%/mr , gdzie :
mr = ms+mw
mr = 200g
Cp = 7%
ms = Cp*mr/100%
ms = 7*200/100 = 14g
mw = mr-ms = 200-14 = 186g
zad.1.
Cm = n/V
V = 0,5dm3
m = 8g i M NaOH = 40g/mol
n = m/M
n = 8/40 = 0,2mol
Cm = 0,2mol/0,5dm3
Cm = 0,4mol/dm3
zad.2.
mr = ms + mw, więc mw = mr - ms
mr = 200g
Cp = 7%
Cp = (ms/mr)*100%, więc ms = (Cp*mr)/100%
ms = (7%*200g)/100% = 14g
mw = 200g - 14g = 186g
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