calculate number of ions present in alcl3 (aluminium chloride) having mass 150g.
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
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