Question

Calculate the molarity of $NaOH$ in the solution prepared by dissolving its $4g$ in water to make a solution of $250ml$.

Answer 1

We know that the molarity is the most generally utilized unit of fixation and is indicated by M. It is characterized as no. of moles of solute present in 1 liter of solution. Consequently,
Formula used to calculate molarity of the solution is,
$Molarity = \dfrac{{{\text{Moles of solute}}}}{{{\text{Volume of solution(L)}}}}$

Complete answer:
We have to remember that the molarity of the solution is calculated by using the formula mentioned in the hint part of the solution.
Given:
The mass of sodium hydroxide is $4g$.
The volume of solution is $250mL$
First, calculate the number of moles of sodium hydroxide,
To calculate the number of moles one must know the molecular mass of the substance. The molecular weight of sodium hydroxide is $40g/mol$.
$Moles\left( {NaOH} \right) = \dfrac{{4g}}{{40g/mol}} = 0.1moles$
Convert the volume of the solution in milliliter to liters,
The volume of the solution is $0.250L$
The molarity of the solution is,
$Molarity = \dfrac{{0.1}}{{0.250}} = 0.4M$
The molarity of sodium hydroxide in the solution prepared by dissolving its $4g$ in water to make a solution of $250mL$ is $0.4M$.

Note:
We need to know that most of the responses occur in solutions thus it is critical to see how the measure of substance is communicated when it is available in the solution. There are numerous manners by which the measure of substances in solution is communicated:
Molality:
We need to remember that the molality (m) is characterized as the quantity of moles of solute per kilogram of dissolvable.
$Molality = \dfrac{{{\text{Moles of solute}}}}{{{\text{Kilograms of dissolvable}}}}$
On the off chance that ninety eight gram of sulphuric acid is available in one kg of water then it is one molal solution of Sulphuric acid in water. Furthermore, consequently molality will be one
Normality:
We have to know that the normality is characterized as the quantity of counterparts per liter of solution:
$Normality = \dfrac{{{\text{Number of counterparts}}}}{{1{\text{ L of arrangement}}}}$