Question

If $2.46{\text{ }}g$ of sodium hydroxide (molar mass = 40) are dissolved in water and the solution is made to $100{\text{ }}c{m^3}$ in a volumetric flask. Calculate the molarity of the solution.
A. $0.555M$
B. $0.516M$
C. $0.615M$
D. $0.715M$

Answer 1

Molarity is defined as the amount of moles of solute present in a litre of solution. It is a concentration term and has the unit called as Molar (M). One Molar is defined as the 1 mole of substance dissolved in 1 litre of the solution and Molar is also having a basic unit of mole/L.

Complete step by step answer:
Let’s start with discussing a little about the concept of molarity for better understanding. Molarity is defined as the amount of moles of solute present in a litre of solution. It is a concentration term and has the unit called as Molar (M). One Molar is defined as the 1 mole of substance dissolved in $1$ litre of the solution and Molar is also having a basic unit of mole/L.
Coming to the question, we are given $2.46{\text{ }}g$ of $NaOH$ having molecular mass of $40{\text{ }}g$ and is dissolved in water and the solution is made up to $100{\text{ }}c{m^3}$. We need to find out the molarity of the solution.
The formula of Molarity is given by
${\text{Molarity = }}\dfrac{{{\text{Moles of Solute (Mole)}}}}{{{\text{Volume of Solution in Litres (L)}}}}$,
We also know that $100{\text{ }}c{m^3}$ is equal to $100{\text{ }}mL$ of solution which is $0.1{\text{ }}L$. Also, the moles of solute can be calculated using ${\text{Moles = }}\dfrac{{{\text{Given Mass}}}}{{{\text{Molar Mass}}}}{\text{ = }}\dfrac{{2.46}}{{40}}{\text{ = 0}}{\text{.0615 moles}}$.
Putting the moles and the volume of solution in the molarity equation we get
${\text{Molarity = }}\dfrac{{0.0615}}{{0.1}}{\text{ = 0}}{\text{.615M}}$

Hence the answer to this question is option C. $0.615{\text{ }}M$

Note:
Molarity as everyone knows is a unit of concentration of solution. One molar solution generally means one mole of solute in one litre of solvent. The concept of molarity is widely used in industry scale as well as in laboratory scale to achieve perfect results. Usually the solutions are being made by weighing the solute and adding them into water. The weighing is done based on the Molarity of the solution require.