Question

# Given, 5.0 g of ${{H}_{2}}{{O}_{2}}$ is present in 100 mL of the solution. The molecular mass of ${{H}_{2}}{{O}_{2}}$ is 34. The molarity of the solution is:A. 1.5 MB. 0.15 MC. 3.0 MD. 50 M

Hint: Recall the formula used to find the molarity of any substance using the number of moles of solute and the volume of solution. Then calculate the number of moles based on the molecular mass and given weight.

We know that molarity is the one of the most widely used terms for the concentration which is denoted by M. We know the formula for the molarity is $\text{molarity = }\dfrac{\text{no}\text{. of moles of solute}}{\text{volume of solution in Litre}}$.
We need to find out the number of moles of solute first. We know the formula for the number of moles. That is,
$\text{No}\text{. of moles = }\dfrac{\text{given mass of solute}}{\text{molecular mass of solute}}$
Given mass of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ = 5 g
Given molecular weight of ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ = 34
We can easily calculate this too if it was not given in the question by multiplying the atomic masses of the elements present by the number of atoms present (of that element) and then add the resulting values. Here it will be $(2\times 1)+(2\times 16)=34$
Therefore, $\text{No}\text{. of moles = }\dfrac{5}{34}=0.147$
Now, the volume of solution is given = 100 mL = 0.1 L
Therefore,
\begin{align} & Molarity=\dfrac{\text{No}\text{. of moles}}{\text{volume of solution}} \\ & Molarity=\dfrac{0.147}{0.1} \\ & Molarity=1.5M \\ \end{align}
So, the molarity of ${{H}_{2}}{{O}_{2}}$ is 1.5 M and

Hence, the correct option for this question is ‘A’.