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प्रश्न
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)
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उत्तर
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
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Complete the following table.
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Acid Radicals `→` |
Chloride |
Nitrate |
Sulphate |
Carbonate |
Hydroxide |
Phosphate |
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Basic Radicals `↓` |
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Magnesium |
MgCl2 |
Mg(NO3)2 |
MgSO4 |
MgCO3 |
Mg(OH)2 |
Mg3(PO4)2 |
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Sodium |
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Zinc |
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Silver |
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Ammonium |
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Calcium |
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Iron (II) |
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Potassium |
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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.
