Question

# When 2.64 g of a hydrated salt $(MS{{O}_{4}}x{{H}_{2}}O)$ is completely dehydrated, 1.20 g of anhydrous salt is obtained. If the molecular weight of anhydrous salt is 120 g $mo{{l}^{-1}}$, what is the value of x ?[A] 2[B] 4[C] 5[D] 6[E] 7

Hint: We have to write the reaction for dehydration of the given salt and as the molecular weight of the anhydrous salt is given to us, we can use that to find the molecular weight of the hydrated salt which will be the same but the molecular weight of the water molecules would be added to it. Following the unitary method, we can use (Molecular weight of anhydrous salt $\times$weight of hydrated salt) $\div$ (molecular weight of hydrated salt) = weight of the anhydrous salt, and solve for ‘x’ to get the answer.

A hydrated salt contains water of crystallization and can be completely dehydrated on heating. On heating, the water part is separated and is evaporated as vapour and the dehydrated salt is left behind as anhydrous salt.
As we can see in the question that the hydrated salt is completely dehydrated therefore, we can write the equation as -
$MS{{O}_{4}}x{{H}_{2}}O\xrightarrow{heat}MS{{O}_{4}}+{{H}_{2}}O$
Where, $MS{{O}_{4}}x{{H}_{2}}O$is the hydrated salt and the anhydrous salt left behind is$MS{{O}_{4}}$.
The weight of the hydrated and the dehydrated salt is given to us which is 2.46 g and 1.20 g respectively, we can write the weights as -
$MS{{O}_{4}}x{{H}_{2}}O\xrightarrow{heat}MS{{O}_{4}}+{{H}_{2}}O$
2.46 g $\to$1.20 g + ‘x’ g
So, let us consider that ‘x’ water molecules were present in the hydrated salt.
The molecular weight of the anhydrous salt, $MS{{O}_{4}}$ is given to us in the question which is 120 g/mol.
Therefore, the molecular weight can be written as 120 + (x$\times$ molecular weight of water), as the hydrated salt contains the same elements as the anhydrous salt; the only added thing is the molecular weight of water. And since we are considering there are ‘x’ moles of water-
Molecular weight of water = (2+16) = 18
Molecular weight of $MS{{O}_{4}}x{{H}_{2}}O$ = 120 + (x$\times$ 18)
Therefore, we can write that 120 + (x$\times$18) g of $MS{{O}_{4}}x{{H}_{2}}O$ gives 120 g of $MS{{O}_{4}}$ on complete dehydration.
Or, following unitary method, we can write that -
1g of $MS{{O}_{4}}x{{H}_{2}}O$ gives 120$\div$[120+(x$\times$ 18)] g of $MS{{O}_{4}}$
Therefore, 2.46 g of $MS{{O}_{4}} x {{H}_{2}}O$ will give (120$\times$2.46)$\div$[120+(x$\times$18)] of $MS{{O}_{4}}$
But according to the question, we will get 1.20 g of $MS{{O}_{4}}$
Therefore, we can write that -
(120$\times$ 2.46)$\div$[120+(x$\times$ 18)] = 1.20
Solving the above equation for x, we get -
x =7.
Hence, there are 7 water molecules in the hydrated salt, which gives us the formula as $MS{{O}_{4}}\cdot 7{{H}_{2}}O$.
Therefore, the correct answer is option [E] 7.

Note: While writing the dehydration equation, we should only break down the water part and not the whole compound. We just need to write down water + remaining part of the compound as this is just a dehydration process i.e. no chemical reagents are added, the salt is just heated on a watch glass till it dries. There are various examples which become anhydrous with heating like Blue Vitriol or Gypsum etc.