[danny411]

calcluate the percent by mass of acetic acid, ch3cooh in a solution of .620M ch3chooh. Density is 1.07 g/ml

molar mass ch3cooh-80.05 g/mol

mass of ch3cooh= (.620m/l)(80.05 g/m)==49.631 g/l

mass of sample 1.07 g/.001==1070 g/l

percent mass ch3cooh= 49.631(g/l)/ 1070(g/l) x100%==4.64%

[Borek]

Calculations look correct, but molar mass used is wrong.