Question

Ile wody należy odparować aby z 200g 3% r-r kwasu borowego otrzymać 30% r-r tego kwasu??

Answer 1

mr₁=200g

Cp₁=3%

Cp₂=30%

 

ms=Cp·mr/100%

ms=3%·200g/100%=6g

 

mr=ms·100%/Cp

mr₂=6g·100%/30%=20g

 

m(rozp.)=mr₁-mr₂

m(rozp.)=200g-20g=180g

 

Odp.: Szukana masa wody, którą należy odparować, wynosi 180g.

Answer 2

mr1 = 200g

Cp1 = 3%

Cp2 = 30%

mw = x

 

Cp = (ms/mr)*100%
ms = (Cp*mr)/100%
ms1 = (200g*3%)/100% = 6g
ms1=ms2
mw = mr1 - ms1

mw = 200g - 6g = 194g

mr = (ms*100%)/Cp
mr2 = (6g*100%)/30%

mr2 = 20g
mw = mr2 - ms1

mw = 20g - 6g = 14g
mw = mw1 - mw2 = 194g - 14g = 180g

pozdrawiam :)