How many mole of chloride ions are present in a 66.7g sample of AlCl3 ?? Need help guys ..... plzz ☺☺☺
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22
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
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
tnwramit1:
tnx
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