Question

Rozwiąż zadania z załącznika

Answer 1

Zad. 9
Zakładam 70cm³ mieszaniny: 50cm³ benzenu, 20cm³ toluenu
Masa benzenu
m= 50cm³*0,87g/cm³ = 43,5g masa toluenu
m =20cm³*0,06g/cm³=1,2g
1,2 g + 43,5g =44,7g
Zawartość benzenu
44,7g-------------100%
43,5g---------------x
X= 43,5g*100%/44,7g =97,3%
Toluen 100%-97,3%=2,7%
Zad. 11
M bromobenzenu 6*12g/mol  + 5*1g/mol + 80g/mol = 157g/mol
M bromonitrobenzenu
 6*12g/mol  + 4*1g/mol + 80g/mol + 14g/mol + 2*16g/mol=202g/mol  
C₆H₅Br  + HNO₃  → C₆H₄Br(NO₂) + H₂O
157g--------------------202g
90g------------------------x
X=90g*202g/157g= 115,8g
115,8g ------------100%
x---------------------67%
x=67%*115,8g/100%=77,6g
zad. 14
ms = cp*mr/100%
ms = 5%*200g/100% = 10g    
C₆H₄(NO₂)₂
M = 12g/mol*6 + 1g/mol*4+ 14g/mol*2 + 4*16g/mol =168g/mol   168g------------1mol
10g---------------x
X = 10g*1mol/169g =0,06moli