Questions & Answers

Question

Answers

[A] 2

[B] 4

[C] 5

[D] 6

[E] 7

Answer
Verified

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 $MS{{O}_{4}}\cdot 7{{H}_{2}}O$.