Question

For $1\% \left( {w/v} \right)$ of an aqueous solution of $NaOH$ is isomolar with:
A.$1M$, ${H_2}S{O_4}$
B.$0.1M$, $NaCl$
C.$0.5M$, $KOH$
D.$0.25M$, $HCl$

Answer 1

Hint: This question gives the knowledge about determining the molarity of the solution. Molarity is the ratio of number of moles of solute to the volume of the solution in litres. Molarity is the unit of concentration.
Formula used: The formula used to determine the number of moles is as follows:
$n = \dfrac{w}{m}$
Where $w$ is the given weight, $n$ is the number of moles of solute and $m$ is the molecular weight.
The formula used to determine the molarity of the solution is as follows:
$M = \dfrac{n}{V}$
Where $M$ is the molarity of the solution, $n$ is the number of moles of solute and $V$ is the volume of the solution in litres.

Complete step by step answer:
Molarity is the ratio of number of moles of solute to the volume of the solution in litres. Molarity is the unit of concentration. The SI unit of molarity is $mol/L$ and molarity is generally abbreviated as $M$. Molarity is also termed as molar concentration. Molarity is very helpful in determining the concentration as well as dilution problems.
Now, we will determine the molarity of the solution by using the molarity formula as follows:
$ \Rightarrow M = \dfrac{n}{V}$
According to the question, $1gm$ of sodium hydroxide is present in $100ml$ of solution.
To determine the molarity first we will require the number of moles of solute. The formula used to determine the number of moles is as follows:
$ \Rightarrow n = \dfrac{w}{m}$
Substitute given weight as $1gm$ and molecular weight of $NaOH$ as $40$ in the above formula.
$ \Rightarrow n = \dfrac{1}{{40}}$
Now, to determine the molarity of the solution by substituting $n$ as $\dfrac{1}{{40}}$, volume in litres as $\dfrac{{100}}{{1000}}$ in the molarity formula.
$ \Rightarrow M = \dfrac{1}{{40}} \times \dfrac{{1000}}{{100}}$
On simplifying, we get
$ \Rightarrow M = 0.25$
Therefore, the molarity of the solution is $0.25M$.
Therefore, option $D$ is the correct option.

Note:
Always remember that the molarity is the ratio of number of moles of solute to the volume of the solution in litres. Molarity is used to determine the concentration of the solution. The SI unit of molarity is $mol/L$ and molarity is generally abbreviated as $M$.