Question

The hydrated salt of $N{a_2}C{O_3}.x{H_2}O$ undergoes 63 % loss in mass on heating
and became anhydrous. The value of x is:
(A) 3
(B) 5
(C) 7
(D) 10

Answer 1

Hint: To start answering the above question, we need atomic masses of all elements present, and knowledge of stoichiometry of chemical reactions. We know that heating of hydrated salts will result in loss of water. This loss is given to us in percentage. Now after writing the reaction and using stoichiometry, we can get the value of x easily, by comparing it with the amount of water lost from the given salt.

Complete step by step answer:
First of all, to start answering, let us write chemical reactions.
\[N{a_2}C{O_3}.x{H_2}O\xrightarrow{\Delta }N{a_2}C{O_3} + x{H_2}O\]
Now, let us write the atomic masses of all the elements.
Na = 23 u
C = 12 u
O = 16 u
${H_2}O = 18u$
Now, we can calculate molecular mass of $N{a_2}C{O_3}.x{H_2}O$ = \[23 \times 2 + 12 + 16 \times
3 + 18 \times x\]
On simplification, we get molecular mass of $N{a_2}C{O_3}.x{H_2}O$ = 106 + 18x
From chemical reaction, we observe that the amount of water loss = 18x
Loss in percentage= 63 % (Given in question)
Percentage loss = (amount of water loss/total mass of hydrated salt) $\times $ 100
Now, substitute the values.
\[63 = \dfrac{{18x}}{{106 + 18x}} \times 100\]
Multiplying 100 with 18x, we get
$63 = \dfrac{{1800x}}{{106 + 18x}}$
Taking, the denominator on other side and cross multiplying, we get:
$(63 \times 106) + (63 \times 18x) = 1800x$
Simplify the equation, to get
$6678 + 1134x = 1800x$
Taking x containing values on one side, we get
$6678 = 1800x - 1134x$
$\Rightarrow 6678 = 666x$
Now, we take 666 to divide on other side,
$\dfrac{{6678}}{{666}} = x$
$\therefore x = 10.02 \approx 10$
Now, x is approximately equal to 10.
Substitute value of x in the hydrated formula of salt $N{a_2}C{O_3}.x{H_2}O$
Thus, the new formula of hydrated salt is $N{a_2}C{O_3}.10{H_2}O$
Hence, the correct option is (D) .

Note:
The water that we see in hydrated salt is called water of crystallisation. This even imparts colour to salts. We should not overheat the salt, so much that the salt itself will start dissociation. But we need only loss of water, so heating has to be done under proper care.