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 $N{a_2}S{O_3} \cdot 4{H_2}O$,

& {\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.