Question

Oblicz ile moli tlenku siarki VI znajduje się w 250cm³ 10% roztworu, o gęstości 1,048 g/cm³.

Answer 1

d=1,048g/cm3 = 1048g/dm3

V=250cm3= 0,25dm3

 

mr=d*V

mr=1048*0,25 = 262g

 

Cp=ms*100%/mr

10%=ms*100%/262  /*262

2620=ms*100   /:100

ms= 26,2g

 

MmolowaS03= 80g/mol

m=M*n  /:M

n=m/M

n=26,2/80

n=0,33 mola

Answer 2

v=250cm3

d=1,048g/cm3

m=v*d

m=250*1,048

m = 262g

 

Cp=10%

mr=262g

ms=Cp*mr/100%

ms=10*262/100

ms = 26,2g

 

M SO3=80g/mol

80g SO3 ----- 1mol

26,2g SO3 ----- xmoli

x = 0,33mola