Question

The hydrated salt $N{a_2}C{O_3}.x{H_2}O$ undergoes 46% loss in mass on heating and becomes anhydrous. The value of “x” is

Answer 1

Hint: We know that when certain substances are exposed to water present in the atmosphere. They either absorb it on their surface or have it in their structure producing a complex where water bonds with the cation in ionic substances. The water present in the latter case is known as water of crystallization (or) water of hydration.

Complete step by step answer:
We can define hydrates are inorganic salts comprising molecules of water combined in a fixed ratio as an integral part of the crystal which are either attached to the center of metal or that has crystallized with the metal complex. Such hydrates are also said to have water of crystallization or water of hydration.
Given data contains,
Mass percentage of water lost is \[46\% \].
The molar mass of sodium carbonate is $106g/mol$.
The molar mass of water is $18g/mol$.
We can consider the mass of the salt is W.
Mass of salt is the sum of the molar mass of sodium carbonate and the molar mass of water.
Then, $106 + 18x = W \to \left( 1 \right)$
We know that 46% of water molecules are lost and 54% is the rest of the salt mass percentage.
After heating, $\dfrac{{54W}}{{100}} = 106 \to \left( 2 \right)$
We have to equation (2) in (1),
$106 + 18x = \dfrac{{106 \times 100}}{{54}}$
On simplifying we get,
$ \Rightarrow 18x = 90$
$ \Rightarrow x = \dfrac{{90}}{{18}}$
$ \Rightarrow x = 5$
The value of x is 5.

The value of x in $N{a_2}C{O_3}.x{H_2}O$ is $N{a_2}C{O_3}.5{H_2}O$.

Note: We have to remember that a hydrate which has lost a molecule of water is known as an anhydride. We could remove the remaining water with very strong heating. Anhydrous is the substance that does not contain any water molecules. Few anhydrous compounds are hydrated very easily that they are hygroscopic in nature. We can use them as drying agents or desiccants.