Question

A hydrate of $$N{a_2}S{O_3}$$ has 50% water by mass. It is……
(A) $$N{a_2}S{O_3} \cdot 4{H_2}O$$
(B) $$N{a_2}S{O_3} \cdot 6{H_2}O$$
(C) $$N{a_2}S{O_3} \cdot 7{H_2}O$$
(D) $$N{a_2}S{O_3} \cdot 2{H_2}O$$

Answer 1

Hint: Atomic Weight of Sulfur atom is 32$$gmmo{l^{ - 1}}$$. We can find % weight of water in all the given compounds after calculating the molecular weight of all of them. Then we can match % weight of the water content and find the answer.

Complete step by step solution:
Let’s calculate molecular weight of all the compounds given in the options and then we will find their % composition of water content
For $$N{a_2}S{O_3} \cdot 4{H_2}O$$,
$$\eqalign{
  & {\text{Molecular weight of N}}{{\text{a}}_2}S{O_3} \cdot 4{H_2}O = {\text{2(Atomic weight of Na) + Atomic weight of sulfur}} \cr
  & {\text{ + 3(Atomic weight of oxygen) + 4(Molecular weight of water)}} \cr} $$
Molecular Weight of $$N{a_2}S{O_3} \cdot 4{H_2}O$$= 2(23) + 32 + 3(16) + 4(18)
Molecular Weight of $$N{a_2}S{O_3} \cdot 4{H_2}O$$= 198$$gmmo{l^{ - 1}}$$

Now,
$$\eqalign{
  & {\text{Molecular weight of N}}{{\text{a}}_2}S{O_3} \cdot 6{H_2}O = {\text{2(Atomic weight of Na) + Atomic weight of sulfur}} \cr
  & {\text{ + 3(Atomic weight of oxygen) + 6(Molecular weight of water)}} \cr} $$
Molecular weight of $$N{a_2}S{O_3} \cdot 6{H_2}O$$ = 2(23) + 32 + 3(16) + 6(18)
Molecular weight of $$N{a_2}S{O_3} \cdot 6{H_2}O$$ = 234$$gmmo{l^{ - 1}}$$
$$\eqalign{
  & {\text{Molecular weight of N}}{{\text{a}}_2}S{O_3} \cdot 7{H_2}O = {\text{2(Atomic weight of Na) + Atomic weight of sulfur}} \cr
  & {\text{ + 3(Atomic weight of oxygen) + 7(Molecular weight of water)}} \cr} $$
Molecular weight of $$N{a_2}S{O_3} \cdot 7{H_2}O$$ = 2(23) + 32 + 3(16) + 7(18)
Molecular weight of $$N{a_2}S{O_3} \cdot 7{H_2}O$$ = 252 $$gmmo{l^{ - 1}}$$

Also
$$\eqalign{
  & {\text{Molecular weight of N}}{{\text{a}}_2}S{O_3} \cdot 2{H_2}O = {\text{2(Atomic weight of Na) + Atomic weight of sulfur}} \cr
  & {\text{ + 3(Atomic weight of oxygen) + 2(Molecular weight of water)}} \cr} $$
Molecular weight of $$N{a_2}S{O_3} \cdot 2{H_2}O$$= 2(23) + 32 + 3(16) + 2(18)
Molecular weight of $$N{a_2}S{O_3} \cdot 2{H_2}O$$= 162$$gmmo{l^{ - 1}}$$

Now , we can find % water content in the compound by following the formula.
$$Water\% = \frac{{{\text{100}} \times {\text{Mass of water in molecule }}}}{{{\text{Mass of molecule}}}}$$
Now, For $$N{a_2}S{O_3} \cdot 4{H_2}O$$,
$$Water\% = \frac{{100 \times 72}}{{198}}$$
$$Water\% = 36.36\% $$
For $$N{a_2}S{O_3} \cdot 6{H_2}O$$,
$$Water\% = \frac{{100 \times 108}}{{234}}$$
$$Water\% = 46.15\% $$
For $$N{a_2}S{O_3} \cdot 7{H_2}O$$
$$Water\% = \frac{{100 \times 126}}{{252}}$$
$$Water\% = 50\% $$
And for $$N{a_2}S{O_3} \cdot 2{H_2}O$$
$$Water\% = \frac{{100 \times 36}}{{162}}$$
$$Water\% = 22.22\% $$

Based upon water content by % derived as above, we can say that $$N{a_2}S{O_3} \cdot 7{H_2}O$$ has Water by 50% of its mass.
So, Correct option is (C) $$N{a_2}S{O_3} \cdot 7{H_2}O$$

- Atomic weight of Sulphur = 32$$gmmo{l^{ - 1}}$$
- Atomic Weight of Sodium = 23$$gmmo{l^{ - 1}}$$

- Alternative Method:
We can simply calculate mass of $$N{a_2}S{O_3}$$ which is equal to 126. Now as water content in the molecule is required to be 50% of the total molecular mass, total weight of water in the molecule needs to be equal to 126.
In case of $$N{a_2}S{O_3} \cdot 7{H_2}O$$ only, Total water mass is equal to 126.
So, in this manner we can also get the answer. This is a short trick.

Note: Consider all the water molecules present in the molecule in order to find % water content in the molecule. As shown in the alternative method, we got the answer directly but even in those cases check the % content by applying % formula also.