Question 11 Calculate the number of aluminium ions present in 0.051 g 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 = 27 u)
Class 9 - Science - Atoms and Molecules Page 44
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Answered by
117
Heya !
Here is youre answer ===========================================================================
>> This problem involves aluminium ions. please note that the mass of aluminium ion is the same as that of an aluminium atom. In order to solve this problem, first of all we have to find out the mass of aluminium atoms in 0.051 g of aluminum oxide (Which will give us the mass of aluminium ions) This can be done as follows:
1 mole of Al2O3 = Formula mass of Al2O3 in grams
= Mass of Al × 2 + Mass of O × 3
= 27 × 2 + 16 × 3
= 54 + 48
= 102 grams
Now, 1 mole of Al2O3 contains 2 moles of Al.
So, Mass of Al in 1 mole of Al2O3 = Mass of Al × 2
= 27 × 2
= 54 grams
Now, 102 g aluminum oxide contains = 54 g Al
So 0.051 g aluminum oxide contains = 54/ 102 × 0.051 g Al
= 0.027 g Al
The atomic mass of aluminium is given to be 27 u. This means that 1 mole of aluminium atoms ( or aluminium ions ) has mass of 27 grams , and it contains 6.022 × 10^23 aluminium ions.
Now, 27 g of aluminium has ions = 6.022 × 10^23 aluminium ions.
So, 0.027 g of aluminium has ions = 6.022 × 10^23 / 27 × 0.027.
= 6.022 × 10^20
Thus, the number of aluminium ions (Al³+) in 0.051 gram of aluminium oxide is 6.022 × 10^20.
Hope it helps !!!
Here is youre answer ===========================================================================
>> This problem involves aluminium ions. please note that the mass of aluminium ion is the same as that of an aluminium atom. In order to solve this problem, first of all we have to find out the mass of aluminium atoms in 0.051 g of aluminum oxide (Which will give us the mass of aluminium ions) This can be done as follows:
1 mole of Al2O3 = Formula mass of Al2O3 in grams
= Mass of Al × 2 + Mass of O × 3
= 27 × 2 + 16 × 3
= 54 + 48
= 102 grams
Now, 1 mole of Al2O3 contains 2 moles of Al.
So, Mass of Al in 1 mole of Al2O3 = Mass of Al × 2
= 27 × 2
= 54 grams
Now, 102 g aluminum oxide contains = 54 g Al
So 0.051 g aluminum oxide contains = 54/ 102 × 0.051 g Al
= 0.027 g Al
The atomic mass of aluminium is given to be 27 u. This means that 1 mole of aluminium atoms ( or aluminium ions ) has mass of 27 grams , and it contains 6.022 × 10^23 aluminium ions.
Now, 27 g of aluminium has ions = 6.022 × 10^23 aluminium ions.
So, 0.027 g of aluminium has ions = 6.022 × 10^23 / 27 × 0.027.
= 6.022 × 10^20
Thus, the number of aluminium ions (Al³+) in 0.051 gram of aluminium oxide is 6.022 × 10^20.
Hope it helps !!!
Answered by
89
Hi !
Aluminum oxide = Al₂O₃
Atomic mass of Al = 27
Atomic mass of Oxygen = 16
There are two atoms of aluminium , hence ,
atomic mass = 27 x 2 = 54
three atoms of oxygen are there , hence ,
atomic mass = 16 x 3 = 48
atomic mass / molar mass of aluminium oxide = 54 + 48 = 102 g
Given mass = 0.051 g
No: of moles = given mass / molar mass
= 0.051 /102 moles
we have a formula to calculate no: of particles in an atom .
The no: of particles can be ions , molecules , etc.
No: of particles = No: of moles x Avogadro no:
As there are 2 atoms of aluminium in aluminium oxide , we multiply the product obtained above with 2.
No: of ions = 0.051/102 x 6.022 x 10²³ x 2
= 6.022 × 10 ²⁰ aluminium ions
Aluminum oxide = Al₂O₃
Atomic mass of Al = 27
Atomic mass of Oxygen = 16
There are two atoms of aluminium , hence ,
atomic mass = 27 x 2 = 54
three atoms of oxygen are there , hence ,
atomic mass = 16 x 3 = 48
atomic mass / molar mass of aluminium oxide = 54 + 48 = 102 g
Given mass = 0.051 g
No: of moles = given mass / molar mass
= 0.051 /102 moles
we have a formula to calculate no: of particles in an atom .
The no: of particles can be ions , molecules , etc.
No: of particles = No: of moles x Avogadro no:
As there are 2 atoms of aluminium in aluminium oxide , we multiply the product obtained above with 2.
No: of ions = 0.051/102 x 6.022 x 10²³ x 2
= 6.022 × 10 ²⁰ aluminium ions
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