Question

The strengths of 5.6 volume of hydrogen peroxide (of density 1 g/ml) in terms of mass percentage and molarity (M), respectively, are:
(Take the molar mass of hydrogen peroxide as 34 g/mol)
A) 0.85 and 0.5
B) 0.85 and 0.25
C) 1.7 and 0.25
D) 1.7 and 0.5

Answer 1

Hint: Since we have the volume strength and density of hydrogen peroxide. First, we can calculate the molarity by dividing volume strength by 11.2 litres. Before calculation, we need to find out the mass of the solution (in grams) by the given density.

Formula used:
The molarity of a substance can be calculated by the given formula;
\[Volume\,strength = 11.2 \times molarity\]
In order to calculate the total mass of the solution, the given formula is to be used:
\[mass\,of\,solution = 1000ml \times density\]
To calculate the mass percentage of hydrogen peroxide we need to find out the mass of solute, which can be done by:
\[mass\,of\,solute = moles \times molar\,mass\]

Complete Step by Step Solution:
It is given that; the volume strength of hydrogen peroxide is 5.6
The density of hydrogen peroxide is 1g/ml, therefore the molarity is given by
\[Volume\,strength = 11.2 \times molarity\]
\[Molarity = \dfrac{{5.6}}{{11.2}} = 0.5\]
Assuming that the solution is of 1 liter,
\[mass\,of\,solution = 1000ml \times density\]
\[mass\,of\,solution = 1000ml \times 1g/ml\]
\[ = 1000g\]
Now, \[mass\,of\,solute = moles \times molar\,mass\], therefore
\[ = 0.5mol \times 34g = 17gm\]
The mass percentage of hydrogen peroxide is given by,
Mass percentage = \[\dfrac{{17}}{{1000}} \times 100 = 1.7\% \]
Hence, option D is the correct answer

Note: Hydrogen peroxide also offers a wide variety of uses such as it is used for the treatment of minor cuts, as an antiseptic agent, for rinsing of the mouth to remove mucous, and also helps in the treatment of minor mouth irritations such as cold sores and cankers.