Question

The hydrated salt ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{.x}}{{\text{H}}_{\text{2}}}{\text{ O}}$ undergoes $63\%$ loss in mass on heating mad become anhydrous. The value of $x$ is:
(Gram atomic mass: ${\text{Na}}$= $23$, ${\text{C}}$=$12$, ${\text{O}}$-$16$)
A. $10$
B.$12$
C.$8$
D.$18$

Answer 1

Find total mass will be the mass of x water molecules and one sodium carbonate molecule. By substituting the mass of anhydrous sodium carbonate, the number of water molecules can be
determined.

Complete Step by step answer: Total mass is equal to $100\%$. Out of which $63\%$ is of water molecules and remaining $37\%$ is of anhydrous sodium carbonate.
Mass of sodium carbonate is determined by the sum of atomic mass of sodium, carbon and oxygen as follows:
$= \left( {23\,\, \times \,2} \right)\, + \,\left( {12 \times \,1} \right)\,\, + \left( {16\, \times \,3} \right)\,$.
Mass of sodium carbonate is ${\text{ = 106}}\,{\text{g/mol}}$
Before heating the mass of salt (hydrated sodium carbonate ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{.x}}{{\text{H}}_{\text{2}}}{\text{O}}$) is as follows:
${\text{Mass }}\,{\text{of }}\,{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{.x}}{{\text{H}}_{\text{2}}}{\text{O}}\,\,{\text{ = }}\,\,{\text{106 + 18}}x$ …… $(1)$
After heating all the water molecules escape so, the mass of salt (anhydrous sodium carbonate ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$) is as follows:
${\text{Mass }}\,{\text{of }}\,{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}\,\,{\text{ = }}\,\,{\text{106}}\,{\text{g/mol}}$ …… $(2)$
After heating the mass of the salt is equal to means $100\,{\text{g}}$ hydrated salt contain $37\,{\text{g}}$ anhydrous sodium carbonate so, the amount of hydrated salt in $106\,{\text{g}}$ anhydrous sodium carbonate is,
$37\,{\text{g N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{ = }}\,100\,{\text{g N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{.x}}{{\text{H}}_{\text{2}}}{\text{O}}\,\,$
$\,{\text{mass of N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{.x}}{{\text{H}}_{\text{2}}}{\text{O}}\,\, = \,\dfrac{{\,106\, \times 100}}{{37}}$
$\,{\text{mass of N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{.x}}{{\text{H}}_{\text{2}}}{\text{O}}\,\, = \,286.48\,{\text{g}}$
So, the mass of hydrate sodium carbonate is $286.48\,{\text{g}}$
Substitute the values from equation $(2)$ into equation $(1)$.
$\,{\text{106}}\,{\text{g}}\,{\text{ + 18}}x\,\,{\text{g}} = \,286.48\,{\text{g}}$
${\text{18}}x\,\,{\text{g}} = \,286.48\,{\text{g}}\, - {\text{106}}\,{\text{g}}$
$x\, = \,\dfrac{{180.48\,{\text{g}}}}{{18\,{\text{g}}}}$
$\,x\, = 10.02$
So, the value of $x$ that is the number of the water molecules is$10$. So, option (A) is correct.
The number of the water molecules is not $12$. So, option (B) is incorrect.
The number of the water molecules is not $8$. So, option (C) is incorrect.
The number of the water molecules is not $18$. So, option (D) is incorrect.

So, option (A) is correct.

Note: Total mass of a compound is calculated by adding the mass of each atom. As the number of a species cannot be fractional so, convert the fraction into an integer. We should keep in mind that while calculating the molecular mass of the compound the value of atomic mass of each element should be taken and should be aware about the isotopes involved.