Question

One gram of alum, KAl(SO4)2.12H2O, contains 1.3 × 1021 Al atoms. How many oxygen atoms are contained in 1.0 g alum?

Answer 1

From the chemical formula, 1 formula unit of KAl (SO4)2.12H2O encompasses 1 atom of Al = 4 * 2 atoms of O in KAl (SO4)2 + 12 atoms of O in 12H2O which is equal to 20 atoms of O.

So, if you have 1.3 × 10^21 Al atoms, you have 20 * 1.3 × 10^21 O atoms will now be equal to 2.6 * 10^22 atoms of O.

Answer 2

Answer: The number of oxygen atoms in given amount of alum are $2.53\times 10^{22}$

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of alum = 1 g

Molar mass of alum = 477.3884 g/mol

Putting values in above equation, we get:

$\text{Moles of alum}=\frac{1g}{477.3884g/mol}=0.0021mol$

In 1 mole of $KAl(SO_4)_2.12H_2O$, 1 mole of aluminium atom, 20 moles of oxygen atoms, 1 mole of potassium atom, 2 moles of sulfur atom and 21 moles of hydrogen atoms are present.

According to the mole concept:

1 mole of a substance contains $6.022\times 10^{23}$ number of atoms.

So, 0.0021 moles of alum will contain $20\times 0.0021\times 6.022\times 10^{23}=2.53\times 10^{22}$ number of oxygen atoms.

Hence, the number of oxygen atoms in given amount of alum are $2.53\times 10^{22}$