Question

How many moles are in \[98.3\] formula units of aluminum hydroxide $Al{{(OH)}_{3}}$?

Answer 1

Hint: This question is based on the concept of number of moles. A formula unit indicates the lowest whole number ratio of ions in any compound. In this question $98.3$formula units simply mean$98.3$g of aluminum hydroxide. In order to find the required number of moles we will firstly find the molar mass of aluminum hydroxide $Al{{\left( O{{H}_{{}}} \right)}_{3}}$. After this we can find the number of moles by dividing the given mass of aluminum hydroxide to its molar mass. This is basically represented by the formula as mentioned below
Number of moles = $\dfrac{given\,mass}{molar\,mass}$

Complete solution:
Firstly, we need to find the molar mass of aluminum hydroxide.
For this we should know the molar masses of few basic elements,
Molar mass of $Al\,=\,27\,$g/mol
Molar mass of $O=\,16$g/mol
Molar mass of H=$1$ g/mol
Therefore, molar mass of $Al{{\left( OH \right)}_{3\,}}\,=\,26\,+\,3\,\left( 16\,+\,1 \right)\,=\,77$g/mol
Also given mass of aluminum hydroxide in the question= $98.3$g
Now using the formula;
Number of moles =$\dfrac{given\,mass}{molar\,mass}$
$\Rightarrow \,\dfrac{98.3}{77}\,=\,1.27$moles
Hence, $1.27$moles of $Al{{\left( OH \right)}_{3}}$is present in a $98.3$ formula unit of aluminum hydroxide.

Note:In such type of questions always find the molar masses of the compounds carefully. Try to learn the molar masses of all the basic elements. Always find out the starting and ending point in order to figure out the total number of conversions necessary for the solution. Always use the basic rules of significant figures in order to express your decimal result in the nearest approximate value. Perform the mathematical operations carefully.