Question

The crystalline salt, on heating, loses$55.9{\text{ }}\%$of its weight. The formula of the crystalline salt is
A) $N{a_2}S{O_4}.5{\text{ }}{H_2}O$
B) $N{a_2}S{O_4}.7{H_2}O$
C) $N{a_2}S{O_4}.10{H_2}O$
D) $N{a_2}S{O_4}.2{H_2}O$

Answer 1

In this type questions, firstly we find the atomic mass of all elements present, and knowledge of stoichiometry of chemical reaction. We know that by heating hydrated salt it loses water. This loss given to us in percentage, we will write the reaction and using stoichiometry then we get the molecules of water, by comparing it with the amount of water lost in from given salt.

Complete step-by-step answer:
The crystalline salt on heating, We got$N{a_2}S{O_4}.x{H_2}O$ where $x$, is the number of water of crystallization to be determined.

Firstly, we write the dehydration reaction as follows....
N{a_2}S{O_4}.x{H_2}O$$$$\xrightarrow{\Delta }\;N{a_2}S{o_4}{\text{ }} + {\text{ }}{H_2}0
Now, let us write the atomic masses of all elements.
$Na = {\text{ }}23$
$C = 12{\text{ }}$
${\text{O}} = 16$
${H_2}O = 18$
We can calculate the molecular mass of the –
$N{a_2}S{O_4}.x{H_2}O$
$= \left( {2{\text{ }} \times {\text{ }}23} \right) + 32 + \left( {16{\text{ }} \times {\text{ }}4} \right) + x \times 18$
$= 142 + 18x{\text{ }}g/m$
For the chemical reaction, we observe that the amount of water loss = $18x$
Loss in percentage (as given in question) = $55.9\%$
%percentage loss = $\left( {{\text{ }} \dfrac{{amount{\text{ }}of{\text{ }}water{\text{ }}loss}}{{total{\text{ }}mass{\text{ }}of{\text{ }}hydrated{\text{ }}salt}}} \right){\text{ }} \times {\text{100}}$
Now, putting the values as per formulae,
$55.9{\text{ }}\% {\text{ }}water = \left( { \dfrac{{18x}}{{142}} + {\text{ }}18x} \right) \times 100$
Multiplying by hundred with $\left( {18x} \right)$
$55.9{\text{ }} = \dfrac{{1800}}{{142 + 18x}}{\text{ }}$
Take the denominator on other side and cross multiplying,
we get,
$\Rightarrow \left( {55.9 \times 142} \right) + \left( {55.9 \times 18x{\text{ }}} \right) = 1800$
$\Rightarrow 7937.8 = 1800{\text{ }} - 1006.2x$
$\Rightarrow 7937.8 = 793.8x$
$x = 9.99 \approx 10{\text{ }}\left( {approx.} \right)$
Substitute this value in given hydrated salt-
$N{a_2}S{O_4}.x{H_2}O$
Now, the molecular formula of compound is -
$N{a_2}S{O_4}.10{H_2}O$

Hence the correct answer is option (C)

Note: The water in hydrated salt is called water of crystallization. A salt with associated water of crystallization is known as a hydrate. The structure of hydrates can be quite elaborate, because of the existence of hydrogen bonds that define polymeric structures. This even impart colour to the salt. We should not overheat the salt, so much that the salt will start dissociation.