Question

calculate number of ions present in alcl3 (aluminium chloride) having mass 150g.

Answer 1

Molecular Mass of AlCl3 is (27*1)+(35.5*3)=133.5 amu

In 133.5 amu AlCl3

Mass of aluminium  =27

Mass of chlorine =106.5

So in 150 amu AlCl3

Mass of aluminium=30.33

Mass of chlorine=119.66

No . Of moles of Al present in AlCl3= 30.33/27=1.12 moles

No. Of moles of Cl present in AlCl3=119.66/35.5=3.37 moles

Therefore no. Of aluminium atoms presesnt in AlCl3=1.12 * Avogadro no.

=6.74*10^23 atoms

No. Of chloride atoms present in AlCl3= 3.37* Avogadro no.

=2.02 *10^24 atoms