1.

calculate the total no. of ions present in 0.051 g of Al2O3. do step wise

Answer»

Let us compute the molar mass of the molecule first

Mass of Al = 27 AMUMass of O = 16 AMUSo mass of Al2O3 "molecule" = 2x27 + 3x16 = 102 AMU

So 1 mole of ALUMINIUM OXIDE will weigh 102 g1 mole of any pure substance contains Avogrado's number of molecules, that is 6.023x10^23.Thus 0.051g of ALUMINIUM OXIDE will contain 6.023x(10^23)x.051/102 = 6.023x(10^20)/2 "molecules "

Each molecule contains two ions of aluminium.

So total no. Of ions in the given amount of substance = 2x6.023x(10^20)/2 =6.023x10^20


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