To find : the number of moles and molecules present in 22 g of acetic acid
Solution :
molecular mass of CH3COOH =(2×atomic mass of C)+(4× atomic mass of H)+(2×atomic mass of O)
=(2× 12u)+(4×1u)+(2×16u)
=60u
Number of moles =\(mass\over molar \space mass\)
=\(22\over 60\)
=0.3667 mol
Number of molecules = Number of moles × Avogadro's constant
=0.3667 × 6.022×1023 molecules
=2.208×1023 molecules