Question

How do you calculate the number of moles of acetic acid present in the aliquot (\[25mL\] of a \[10\% \] vinegar solution)?

Answer 1

Hint: To find moles in a given solution. Firstly, find the mass (in grams) of the solution given in the question, then you can calculate the molar mass and can get the number of moles required.

Complete step by step answer:
Soln.- Soln. is a homogeneous mixture of solute of dissolved into a solvent
Solute- a solute substance that can be dissolved by a solvent to create a solution
Solvent- A substance, capable of dissolving another substance
Solution percent are expressed in the form of\[weight/volume{\text{ }}\% ,{\text{ }}weight/weight{\text{ }}\% ,{\text{ }}or{\text{ }}volume/volume{\text{ }}\% .\] the amount of acetic acid present in \[100\] ml of solution
In vinegar, the water is the solvent and acetic acid is the solute
Assuming that the given solution is \[10\% \] (W/V) vinegar solution. This is \[10\] g of Acetic acid in \[100{\text{ }}mL\] of solution.
Aliquot is=\[25mL\] = There are $\dfrac{{25}}{{100}}$ \[x{\text{ }}10\] =\[2.5{\text{ }}g\] of acetic acid.
Molecular formula of acetic acid \[C{H_3}COOH\]
Its Molar mass using average atomic masses of its constituents is =\[2 \times 12.001\] \[ + {\text{ }}4 \times 1.0079\] \[ + {\text{ }}2 \times 15.9994\] \[\]\[ = 60.03{\text{ }}g \cdot mol - 1\]
Therefore, Number of moles of acetic acid present in the aliquot =$\dfrac{{2.5}}{{60.03}}$
=\[0.0416{\text{ }}mol\]

Note:
The molar mass of compound is equivalent to its atomic mass of constituent atoms
 Write the units ex mol, gm etc.
To calculate molecular mass faster it’s better to learn atomic mass of at least first \[20\] elements
Molar concentration, additionally called molarity, is the quantity of moles of solute per liter of solution.Molarity is the most well-known estimation of solution concentration.