Question

142g of sodium sulphate $N{a_2}S{O_4}$ has:
(A)- 2 moles of $N{a_2}S{O_4}$
(B)- $12.044 \times {10^{23}}$ sodium ions
(C)- 1 mole of $N{a_2}S{O_4}$
(D)- $6.022 \times {10^{23}}$ Sulphur atoms

Answer 1

Hint: The above question is based on mole concept and molar masses. 1 mole of sodium sulphate equals Avogadro's number i.e. $6.022 \times {10^{23}}$. Number of ions of sodium and Sulphur per mole is also equal to Avogadro's number and varies linearly per increase in the number of moles.

Complete step by step solution:
The above question includes a compound named sodium sulphate. Before proceeding further, let's have some basic idea about it.
- Sodium sulfate is the sodium salt of sulfuric acid. Anhydrous sodium sulfate is a white crystalline solid also known as the mineral thernadite, while the decahydrate is known as Glauber’s salt or mirabilis. It is transformed to mirabilite when it is cooled. About two-thirds of the world’s production of sodium sulfate is obtained from mirabilite. It is also produced from by-products of some chemical processes such as hydrochloric acid production.
- Sodium sulfate is electrostatically bonded ionic sulfate. The existence of free sulfate ions in solution is indicated by the easy formation of insoluble sulphates when these solutions are treated with $B{a^{2 + }}$ or $P{b^{2 + }}$ salts. Sodium sulfate is unreactive toward most of the oxidizing and reducing agents. At high temperatures, it can be converted into sodium sulfide by carbothermal reduction (high temperature heating with charcoal etc.).

Now, in order to find which of the options is correct, we have to check each of them.
First, let's find the number of moles of $N{a_2}S{O_4}$.
142g of $N{a_2}S{O_4}$=
$
= (23 \times 2) + 32 + (16 \times 4) \\
= 46 + 32 + 64 \\
= 142g \\
$

Therefore, moles of $N{a_2}S{O_4}$ = $\dfrac{{142}}{{142}}$ = 1 mole of $N{a_2}S{O_4}$.
Now, 1 mole of $N{a_2}S{O_4}$ contains 1 mole of S atoms and 2 moles of Na atoms
Therefore, it contains $6.022 \times {10^{23}}$ atoms of Sulphur
and $2 \times 6.022 \times {10^{23}} = 12.044 \times {10^{23}}$ atoms of Sodium.
So, the correct answer is “Option B”.

Note: Always remember the Avogadro constant ${N_A} = 6.022 \times {10^{23}}$. which was chosen so that the mass of one mole of a chemical compound (in grams) is numerically equal to the average mass of one molecule of the compound.