Question

20 mL of a sample of ${\operatorname{H} _2}{O_2}$ gives 400 mL oxygen measured at \[NTP\]. The sample should be labelled as-
A) \[{\text{5V}}{\text{ }}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\]
B) \[{\text{dil}}{\text{. }}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\]
C) \[{\text{anhy}}{\text{. }}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\]
D) \[{\text{20V }}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\]

Answer 1

Hint: NTP means Normal temperature and pressure, which states that at NTP we can measure the value of volume, mass, pressure, etc. and we get a perfect result, so NTP is Used.
Here given the volume of ${\operatorname{H} _2}{O_2}$ and ${\operatorname{O} _2}$, so the answer may be written in volume. So we have to take a look at option A and option D.

Complete step by step answer:
Here 20 mL of a sample of ${\operatorname{H} _2}{O_2}$ gives 400 mL of oxygen in NTP.
So for labelling, we have to find that 1 mL of ${\operatorname{H} _2}{O_2}$ can give how much volume of oxygen.
So,
20 mL of the sample ${\operatorname{H} _2}{O_2}$ gives 400 mL ${\operatorname{O} _2}$
1 mL of the sample ${\operatorname{H} _2}{O_2}$ gives $\dfrac{{{\text{400}}}}{{{\text{20}}}}{\text{ = 20$\;$mL }}{{\text{O}}_{\text{2}}}$
i.e. The sample should be labelled as $20{\text{V}}\;{H_2}{O_2}$

So the option (D) is correct.

Note: Here we have found the volume percentage sometimes there would be another representation like mole, molar concentration etc.
Then for those cases, we have to solve by calculating mole and molar mass or atomic mass.