Question

Calculate the molarity of 30 percent ($w/v$) $NaOH$ solution, if the density of the solution is $1.05g\,c{{c}^{-1}}$
A. 7.5 M
B. 3.75 M
C. 7.88 M
D. 3.94 M

Answer 1

Hint: Proceed by taking 30 grams of the solute in 100 grams of the solution, and then find out the molarity, using the formula.

Complete answer:
In order to answer the question, we need to learn about the concept of molarity and molality. Both of the terms are used for describing the concentration of a solute in a particular solvent or solution. Now, molality is used when a particular mole of a solute is present in 1kg of the solvent, whereas in case of molarity, the latter part is taken to be 1000 mL of the solution itself.
We also need to learn about another term of concentration that is ($w/v$), which means weight by volume. In the question, we have been given that $NaOH$ is present in 30 percent weight by volume. So, if we take a solution, weighing 100 grams, the amount of $NaOH$ that will be present in it should correspond to 30 grams.
Similarly, if we take 1000 grams , so the amount of $NaOH$that will be present will be equal to 300 grams. This can be done by unitary method, or by multiplying 10, as the ratio will always remain constant. Now, the formula for molarity can be written as:
\[molarity=\dfrac{given\,mass}{molar\,mass}\times \dfrac{1000}{vol\,of\,solution(mL)}\]
We know that the molar mass of the compound $NaOH$ is $40g\,mo{{l}^{-1}}$. So, by substituting the values in the above equation, we get the molarity as:
\[molarity=\dfrac{300}{40}\times \dfrac{1000}{1000}\]
On solving, we get the final molarity as 7.5 M.

This means that the correct answer for our question would be option A.

Note:
It is to be noted that the molarity of a solution can change if the temperature is increased. As $PV=nRT$, so the volume of solution changes, on changing the temperature.