Advertisements
Advertisements
Question
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)
Advertisements
Solution
1 mole of aluminium oxide (Al2O3) = 2 × 27 + 3 × 16 = 102 g
i.e., 102 g of Al2O3 = 6.022 × 1023 molecules of Al2O3
Then, 0.051 g of Al2O3 contains`=(6.022xx10^(23))/102xx0.051" molecules"`
= 3.011 × 1020 molecules of Al2O3
The number of aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.
Therefore, the number of aluminium ions (Al3+) present in 3.011 × 1020 molecules (0.051 g ) of aluminium oxide (Al2O3)
= 2 × 3.011 × 1020
= 6.022 × 1020
RELATED QUESTIONS
Write the names of ZnS compounds :
Fill in the following blank with suitable word:
The particle which is formed by the loss or gain of electrons by an atom is called..............
A particle X has 17 protons, 18 neutrons and 18 electrons. This particle is most likely to be :
Give example.
Trivalent basic radicals
Write the basic radicals and acidic radicals of the following and then write the chemical formulae of these compound.
Nickel bisulphate
Choose the correct answer from the options given below.
The number of carbon atoms in a hydrogen carbonate radical is
Differentiate between the following with suitable examples- A basic radical and an acid radical
Select the correct answer to the statement given below:
An elementary particle of matter – of a group of atoms of different elements behaving as a single unit with charge on the group.
Match the radicals in List I with their correct names from List II.
| List I | List II | |
| 1. `"NO"_3^(1-)` | A: Bisulphite | F: Sulphate |
| 2. `"HSO"_4^(1-)` | B: Phosphate | G: Nitrate |
| 3. `"SO"_4^(2-)` | C: Carbonate | H: Phosphite |
| 4. `"PO"_4^(3-)` | D: Nitrite | I: Sulpite |
| 5. `"HCO"_3^(1-)` | E: Bicarbonate | J: Bisulphate |
Name the following:
The radical containing one nitrogen and four hydrogen atoms.
