find the number of Al ions in 0.051 g of aluminium sulphate

Asked by Devika A | 26th Feb, 2011, 12:00: AM

Expert Answer:

Dear Student
 
1 mole of aluminium sulphate, Al2(SO4)3 = 2 × 27 + 3 ×96 = 342 u = 342 gm

:. 342 gm Al2(SO4)3 has = 6.023 × 1023 Al2(SO4)3 molecules

0.051 gm Al2(SO4)3 has =
6.023 × 1023 × 0.051
= 0.000898 × 1023 molecules
         342

As 1 molecule of Al2(SO4)3 gives = 2Al+++ ions
So, 0.051 gm Al2(SO4)3 gives = 2 × 0.000898 × 1023 ions = 0.001796 × 1023 ions 

We hope that clarifies your query.

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Answered by  | 28th Feb, 2011, 10:23: AM

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