Question

Number of g of oxygen in 32.2 g $$N{a}_{2}S{O}_4.10H_{2}O$$ is:
[Mol.wt = 322]
(A) 16.0
(B)2.24
(C) 18.0
(D) 22.4

Answer 1

Hint: We should know the number of moles of oxygen present in the given 32.2 g of $$N{a}_{2}S{O}_4.10H_{2}O$$ to calculate the number of grams of oxygen present in $$N{a}_{2}S{O}_4.10H_{2}O$$. The formula to calculate the number of moles of oxygen is as follows.
Number of moles of oxygen = $${\displaystyle \frac{\text{weight of the given compound}}{\text{molecular weight of the given compound}}}$$

Complete step by step solution:
-The number of oxygen atoms present in the given sodium sulphate ($$N{a}_{2}S{O}_4.10H_{2}O$$) is 14.
-Molecular weight of the $$N{a}_{2}S{O}_4.10H_{2}O$$ is = 2 (Na) + 1 (S) + 20 (H)+ 14 (O) = 322 g.
-The weight of the sodium sulphate is 32.2 g.
-Therefore the number of moles of oxygen =
$$\begin{array}{rl}& ={\displaystyle \frac{\text{weight of the given compound}}{\text{molecular weight of the given compound}}}\\ & ={\displaystyle \frac{32.2}{322}}\\ & =0.1\end{array}$$
-We know that 14 oxygen atoms present in $$N{a}_{2}S{O}_4.10H_{2}O$$, thus
-Number of moles of oxygen = (0.1) (14) = 1.4.
-32.2 gm of $$N{a}_{2}S{O}_4.10H_{2}O$$contains 1.4 moles of oxygen.
-But we need some grams of oxygen present in 32.2 g of $$N{a}_{2}S{O}_4.10H_{2}O$$.
-To get several grams of the oxygen we have to multiply the number of moles of oxygen with a molecular weight of the oxygen atoms.
-Thus number of grams of oxygen
 $$\begin{array}{rl}& =1.4\times 16\\ & =22.4g\end{array}$$
Therefore Number of g of oxygen in 32.2 g $$N{a}_{2}S{O}_4.10H_{2}O$$ is 22.4 g.

So, the correct option is D.

Note: We have to consider the number of oxygen atoms present in $$N{a}_{2}S{O}_4.10H_{2}O$$ to calculate the number of moles of oxygen. Sometimes we will take 0.1 moles obtained in the initial calculation as the number of moles of oxygen. We are supposed to consider the number of oxygen atoms present in the compound too.