1)ile gramów oraz ile moli kwasu azotowego należy dodac do 1000g H2O aby otrzymać roztwór 40%.
2)ile gramów NaCl musimy dodać do 200g 6% roztworu aby otrzymać roztwor 20%
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1)
Z def. stęż. procent.
40g kwasu-----60g wody (100g roztworu -40g kwasu)
xg kwasu------1000g wody
x = 666,67g kwasu
M HNO3=63g/mol
63g kwasu---------1mol
666,67g kwasu-----xmoli
x = 10,58moli kwasu
2)
xg 100%....................20 - 6 = 14
...............20%
200g 6%..................100 - 20 = 80
x/200 = 14/80
80x = 2800
x = 35g NaCl
zad.1
mw=1000g
Cp=40%
ms=?
Cp=ms/(ms+mw) * 100%
40%=ms/(ms+1000g) * 100% /*ms+1000g
40ms+40000=100ms
60ms=40000 /:60
ms=666,(6) g
M(HNO3)=1g/mol+15g/mol+3*16g/mol=63g/mol
1mol------------------63g
xmol-----------------666,(6)g
x=10,1mol
zad.2
mr₁=200g
Cp₁=6%
ms₁=?
Cp=ms/mr * 100% /*mr
Cp*mr=ms*100% /:100%
ms=Cp*mr/100%
ms₁=6%*200g/100%
ms₁=12g
mw=mr-ms
mw=200g-12g=188g
mw=188g
Cp₂=20%
ms₂=?
Cp=ms/(ms+mw) * 100%
20%=ms/(ms+188g) * 100% /*ms+188g
20ms+3760g=100ms
80ms=3760 /:80
ms=47
masa dosypana=ms₂-ms₁
masa dosypana=35g