1.

How many mole of chloride ions are present in a 66.7g sample of AlCl3 ?? Need help guys ..... plzz ☺☺☺

Answer»

Given

AlCl3

Molar MASS =Al+Cl3

=27+(35X3)

=27+105

=132g

(Given quantity =66.7g)

=66.7/132

=0.5053g

Now TAKE the amount of moles and multiply it by Avogadro's constant, which is 6.023 x 10^23

=0.5053x6.023x10²³

=3.043 x 10²³

This is for 1 ATOM of CL

and in ALCL3 we have 3 atoms


So for three

=3.043x10²³x3

=9.13x10²³

Number of chloride ions=9.13x10²³

This is ur ans hope it will HELP you


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