Question

What is the molarity of $11.2{\text{ V}}$of ${H_2}{O_2}{\text{ ?}}$
$(i){\text{ 1 M}}$
$(ii){\text{ 2 M}}$
$(iii){\text{ 5}}{\text{.6 M}}$
$(iv){\text{ 11}}{\text{.2 M}}$

Answer 1

Hint: We are given with the volume of solution in litres. To find the molarity we need a number of moles of the solute. Thus we will find the number of moles of ${H_2}{O_2}{\text{ }}$. With the help of $n - $ factor we will find the total number of moles of ${H_2}{O_2}{\text{ }}$. Assume all the values at STP.
Formula Used:
$\text{Molarity}{\text{ = }}\dfrac{{{\text{Number of moles of solute}}}}{{{\text{Volume of solution}}}}$

Complete answer:
For finding the molarity of any solution we need a number of moles of solute present in the solution. Number of moles of solute is the ratio of the given mass of solute to the molecular mass of solute. But here in the above equation we are not provided with the mass of ${H_2}{O_2}{\text{ }}$. We are provided with the volume of ${H_2}{O_2}{\text{ }}$ which is equal to $11.2{\text{ V}}$ . Assuming all the conditions at Standard Temperature and Pressure, we can write that:
$1{\text{ L}}$ of ${H_2}{O_2}{\text{ }}$ produces $22.4{\text{ L}}$ of ${O_2}$ at STP
According to question,
$1{\text{ L}}$ of ${H_2}{O_2}{\text{ }}$ produces $11.2{\text{ L}}$ of ${O_2}$ at STP
Therefore the number of moles can be calculated as,
Number of moles $ = {\text{ }}\dfrac{{11.2}}{{22.4}}$
Number of moles $ = {\text{ 0}}{\text{.2}}$
Hence the number of moles of ${H_2}{O_2}{\text{ }}$ is calculated which is equal to $0.2$.
${H_2}{\text{ + }}{{\text{O}}_2}{\text{ }}\xrightarrow{{}}{\text{ }}{{\text{H}}_2}{{\text{O}}_2}$
From the above equation we will find the $n - $ factor for ${H_2}{O_2}{\text{ }}$ .On observing the reaction, we find that the $n - $ factor of the above reaction is $2$. Therefore, the final number of moles of ${H_2}{O_2}{\text{ }}$ is $0.5{\text{ }} \times {\text{ 2 = 1}}$. Now we will find the molarity of solution using the formula,
$\text{Molarity}{\text{ = }}\dfrac{{{\text{Number of moles of solute}}}}{{{\text{Volume of solution}}}}$

On substituting the values we get,
$Molarity{\text{ = }}\dfrac{{1{\text{ mole}}}}{{1{\text{ L}}}}{\text{ = 1M}}$
Thus the molarity of the solution is $1{\text{ }}M$. Hence the correct option is $(i){\text{ 1 M}}$.

Note:
We can find moles of solute by knowing the volume produced at STP. The value of $n - $ factor can be determined by finding the change in oxidation state per molecule. The final number of moles comes after multiplying it with $n - $ factor of solute.