Ile moli kwasu octowego znajduje się w 1dm3 80-procentowego roztworu o gęstości 1.07 g/cm3 ?
d=m/V
m=1000cm3 * 1,07g/cm3
m=1070g roztworu
Cp=ms/mr * 100%
0,8=ms/1070
ms=856g kwasu
60g CH3COOH ---- 1 mol
856g CH3COOH ---- X
X=14,3 mola CH3COOH
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d=m/V
m=1000cm3 * 1,07g/cm3
m=1070g roztworu
Cp=ms/mr * 100%
0,8=ms/1070
ms=856g kwasu
60g CH3COOH ---- 1 mol
856g CH3COOH ---- X
X=14,3 mola CH3COOH