Q. Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.( Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u) Please Explain each and every step!!!Thankx..A lot !!
The formula for aluminium oxide is Al2O3 Its molecular weight is 2*27+3*16= 54+48 =102 Now, 102g of Al2O3 contains 2*6.023*10^23 number of Al ions so 1g contains (2*6.023*10^23)/102 number of Al ions so 0.051g Al2O3 contain (2*6.023*10^23)/102 *0.051 number of Al ions which is equal to 6.023*10^23*10^-3 = 6.023*10^20 number of Al ions(Answer)
