Question

The aluminium sulphate hydrate $\left[ {{\text{A}}{{\text{l}}_{\text{2}}}{{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}}_{\text{3}}}\,{\text{x}}{{\text{H}}_{\text{2}}}{\text{O}}} \right]$ contains 8.20% Al by mass find the no of water molecules associated with each ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_3}$units. If that value is x, find the value of $\dfrac{{\text{x}}}{{\text{3}}}$[i.e write the answer $\dfrac{{\text{x}}}{{\text{3}}}$]

Answer 1

Try calculating the mass percentage of Al in $\left[ {{\text{A}}{{\text{l}}_{\text{2}}}{{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}}_{\text{3}}}\,{\text{x}}{{\text{H}}_{\text{2}}}{\text{O}}} \right]$ aluminium sulphate in terms of x. Find x by equating it to 8.20.

Complete solution:
First we will calculate molar mass of aluminium sulphate i.e $\left[ {{\text{A}}{{\text{l}}_{\text{2}}}{{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}}_{\text{3}}}\,{\text{x}}{{\text{H}}_{\text{2}}}{\text{O}}} \right]$in terms of x.
$\therefore$ Molar mass of aluminium sulphate
= 2x Atomic mass of Al + 3x Atomic mass of S + 12x Atomic mass of ${\text{O}}\,{\text{ + }}\,{\text{2x}}$ (Atomic mass of H) +x$\times$ (Atomic mass of O)
$= \,342\, + \,18{\text{x}}$
$\therefore$ Molar mass of ${\text{A}}{{\text{l}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}\,{\text{x}}{{\text{H}}_{\text{2}}}{\text{O}}\,{\text{ = }}\,{\text{342 + 18x}}$
Now, we will calculate the mass percentage of Al in aluminium sulphate.
{\text{i}}{\text{.e % of Al = }}\,\dfrac{{{\text{Total}}\,{\text{Mass}}\,{\text{of}}\,{\text{Al}}\,{\text{in}}\,{\text{aluminium}}\,{\text{sulphate}}}}{{{\text{Total}}\,{\text{mass}}\,{\text{of}}\,{\text{compound}}\,{\text{aluminium}}\,{\text{sulphate}}}}
But % of Al is given as 8.20% in the question
$\begin{gathered} \Rightarrow \,8.20\, = \,\dfrac{{50 \times \,100}}{{342 + 18{\text{r}}}} \\\ \Rightarrow \,8.20\,(342 + 18{\text{x)}}\,{\text{ = }}\,{\text{5400}} \\\ \Rightarrow \,{\text{8}}{\text{.20}}\, \times \,{\text{342}}\,{\text{ + }}\,{\text{18}}\, \times \,{\text{8}}{\text{.20}}\, \times \,{\text{x = 5400}} \\\ \Rightarrow \,{\text{18}}\, \times \,{\text{8}}{\text{.2x}}\,{\text{ = }}\,{\text{5400}}\,{\text{ - }}\,{\text{2804}}{\text{.4}}\,{\text{ = }}\,{\text{2595}}{\text{.6}} \\\ \Rightarrow \,{\text{x}}\,{\text{ = }}\,{\text{18}} \\\ \end{gathered}$
We got ${\text{x}}\,{\text{ = }}\,{\text{18}}$
$\therefore \,{\text{No}}\,{\text{of}}\,{\text{Water}}\,{\text{molecules}}\,{\text{associated}}\,{\text{ = }}\,{\text{18}}$
Since, we have to find the value of $\dfrac{{\text{x}}}{3}$
We get $\dfrac{x}{3}\, = \,\dfrac{{18}}{3}\, = \,6$.
$\therefore \,\dfrac{{\text{x}}}{3}\, = \,6$ Answer

Note: You need to remember the atomic mass of Al, S, O, H sometimes these one given in the question. You should remember the atomic mass of at least the first 20 elements of the periodic table which you’ll encounter in many problems.