Question

The concentration of ${H_2}{O_2}$ in a solution containing 34 gm in 500ml is given in list I and their corresponding values are given in list II. Then match list I with list II.

List 1	List 2
A.Molarity	1)68
B.Normality	2)224
C. $\% \dfrac{W}{V}$	3)4
D.Volume strength	4)2
5)10

A.(A)-5, (B)-3, (C)-1, (D)-2
B.(A)-4, (B)-3, (C)-1, (D)-2
C.(A)-3, (B)-4, (C)-1, (D)-2
D.(A)-1, (B)-2, (C)-3, (D)-4

Answer 1

We are provided the available weight of hydrogen peroxide in a certain amount of volume. If we know the molar mass of peroxide, we can get the molarity by dividing moles by volume in litres. Once we have the molarity known to us, we can easily determine the normality, $\% \dfrac{W}{V}$ concentration and the volume strength by using their formulas with respect to molarity.

Formulas used: $molarity = \dfrac{{given\,\,mass}}{{molar\,\,mass}} \times \dfrac{1}{{volume(in\,L)}}$ ,
$normality = N \times molarity$ , $volume\,\,strength = 11.2 \times molarity$ and
$\% \dfrac{W}{V} = \dfrac{{given\,\,weight}}{{given\,\,volume(in\,L)}}$

Complete step-by-step answer:
We are provided with a solution containing ${H_2}{O_2}$ of weight 34 gm in 500ml of volume. We know that the molar mass of hydrogen peroxide is 34 grams. We can calculate all the terms given in list I one by one using this information.
First, let us calculate molarity which can be determined by the following formula.
$molarity = \dfrac{{given\,\,mass}}{{molar\,\,mass}} \times \dfrac{1}{{volume(in\,L)}}$
Putting the values in the above formula, as given mass is 34 gm., molar mass is 34 gm, volume is 500ml which is equal to 0.5 litres.
$molarity = \dfrac{{34gm}}{{34gm}} \times \dfrac{1}{{0.5\,L}} = 2M$
Thus, the molarity of hydrogen peroxide in the given solution is 2M. So, A) from list I matches with 4) from list II.
Second is normality of the solution. We can calculate it by the following formula.
$normality = N \times molarity$
Where N is the number of hydrogens it can lose. So, for hydrogen peroxide, we can see it has two hydrogens in its formula and thus, it can lose those two electrons, giving us the value of N = 2.
So, $normality = 2 \times 2M = 4N$
That’s when we say that B) from list I matches with 3) from list II.
Third one is $\% \dfrac{W}{V}$ which can be calculated by the following formula. It is the percentage concentration of weight of solute with respect to the volume of solvent.
$\% \dfrac{W}{V} = \dfrac{{given\,\,weight}}{{given\,\,volume(in\,L)}}$
We know the given mass is 34 gm and given volume is 500ml or 0.5 litres. So, putting these values in it, we get
$\% \dfrac{W}{V} = \dfrac{{34gm}}{{0.5L}} = 68\%$
That’s when we say that C) from list I matches with 1) from list II.
Last one is volume strength of hydrogen peroxide, which is the volume of oxygen released on decomposition of one volume of peroxide at STP. This can be determined by the following formula.
$volume\,\,strength = 11.2 \times molarity$
We know that molarity of solution is 2M, therefore
$volume\,\,strength = 11.2 \times 2M = 22.4$
So, D) from list I matches with 2) from list II.
Hence, the correct option is (B).

Note: We can define normality as the number of equivalents of the solute present in the solution. It is also known as equivalent concentration. Percentage weight in volume is expressed as the number of grams of solute in 100ml of product.