Question

If ${{NaCl}}$ is produced in the equation:
${{FeC}}{{{l}}_{{3}}}{{ + NaOH}} \to {{ Fe(OH}}{{{)}}_{{3}}}{{ + NaCl}}$ was dissolved in water to make a litre of solution, the molarity would be:
1. ${{0}}{{.1M}}$
2. ${{3 M}}$
3. ${{8 M}}$
4. ${{0}}{{.5 M}}$
5. ${{1}}{{.5 M}}$

Answer 1

Molarity (M) is the quantity of a substance in a certain volume of solution. Molarity is defined as moles of solute per litre of solution. It depends on the volume of the solution. The molar concentration of a solution is also known as molarity. ${{NaOH}}$is a strong base.

Complete step by step answer:
The given reaction is as follows.
${{FeC}}{{{l}}_{{3}}}{{ + NaOH}} \to {{ Fe(OH}}{{{)}}_{{3}}}{{ + NaCl}}$
From the reaction, three chlorides ions are replaced by hydroxide ions.
Let’s calculate the number of atoms in the reactant side as well as the product side.

Reactant side	Product side
${{Fe = 1}}$	${{Fe = 1}}$
${{Cl = 3}}$	${{Cl = 1}}$
${{Na = 3}}$	${{Na = 1}}$
${{O = 1}}$	${{O = 3}}$
${{H = 1}}$	${{H = 3}}$

In the reactant side and product side all atoms are equal except sodium and chloride atoms. If we multiply the sodium chloride with three on the product side, all atoms are equal in the product side as well as reactant side.
The equation is as follows.
${{FeC}}{{{l}}_{{3}}}{{ + 3NaOH }} \to {{ Fe(OH}}{{{)}}_{{3}}}{{ + 3NaCl}}$ (balanced equation)
Hence, we can say, three moles of ${{NaOH}}$ are used up to produce three moles of ${{NaCl}}$.
Therefore, the molarity is ${{3M}}$.
Hence, the correct option is (2).

Additional Information:
Reaction of sodium hydroxide and acid produces water and ionic compound.
Sodium hydroxide is extremely basic and reacts with other acids that respond to neutralisation.

$${{NaOH + HCl}} \to {{NaCl + }}{{{H}}_{{2}}}{{O}} \\\ {{NaOH + }}{{{H}}_{{2}}}{{S}}{{{O}}_{{4}}} \to {{NaHS}}{{{O}}_{{4}}}{{ + }}{{{H}}_{{2}}}{{O}} \\\$$

Note:
There are some rules to particularly follow to balance the chemical equations. There are adjustments to the coefficients which are Infront of the reactants or product but do not change the subscripts of the products and reactants. And also consider the polyatomic ions as a whole one.