Fewpal
+1 vote
2.1k views
in Chemistry by (29.7k points)

Calculate the number of aluminium ions present in 0.051g of aluminium oxide. (Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27u)

1 Answer

+2 votes
by (127k points)
selected by
 
Best answer

1 mole of aluminium oxide (Al2O3) = 2 × 27 + 3 × 16 = 102g 

i.e., 102g of Al2O3 = 6.022 × 1023 molecules of Al2O

Then, 0.051 g of Al2O3 contains = (6.022×1023 /102) × 0.051 molecules 

= 3.011 × 1020 molecules of Al2O3 

The number of aluminium ions (Al3+ ) present in one molecules of aluminium oxide is 2. 

Therefore, The number of aluminium ions (Al3+ ) present in 

3.11 × 1020 molecules (0.051g) of aluminium oxide (Al2O3) = 2 × 3.011 × 1020 = 6.022 × 1020

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...