How many moles of acetic acid, ch3cooh, are contained within 24.71 g of
acetic acid?

0.4118 mol

Explanation:

molar mass of CH3COOH = 12*2 + 4*1 +16*2 = 24+4+32= 60 g/mol

24.71 g / 60 (g/mol) = 0.4118 mol

0.41 moles of acetic acid

Explanation:

moles are calculated as the division of compound mass by compound molar weight, that is:

moles of acetic acid = mass of acetic acid/ molar weight of acetic acid

The mass of acetic acid is 24.71 g  and its molar weight is 2*12 + 4*1+ 2*16 = 60 g/mol, so

moles of acetic acid = 24.71/60 = 0.41 moles

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