Odpowiedzi

2011-07-25T14:40:45+02:00

M(C₆H₁₂O₆)=6*12g+12*1g+6*1=180g

M(10H₂O)=10*(2*1g+16g)=180g

 

ms=180g

mr=180g+180g=360g

Cp=?

 

Cp=ms/mr * 100%

Cp=180g/360g * 100%

Cp=50%

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2011-07-25T14:41:21+02:00

1 mol C₆H₁₂O₆ = 180g

10 moli H₂O = 18g·10=180g

 

ms=180g

mr=2·180g=340g

 

Cp=\frac{ms}{mr}\cdot100\ \%

Cp=\frac{180g}{340g}\cdot100\ \%

Cp=\frac{1}{2}\cdot100\ \%

Cp=50\ \%

 

Odp: Stężenie procentowe roztworu glukozy wynosi 50%.

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