Question

The sample of ${{H}_{2}}{{O}_{2}}$ solution labeled as 28 volume has density of 26.5 g/L. mark the correct option(s) representing the concentration of same solution in other units.
(A) ${{M}_{{{H}_{2}}{{O}_{2}}}}=2.5$
(B) $%\dfrac{w}{v}=17$
(C) mole fraction of ${{H}_{2}}{{O}_{2}}$= 0.2
(D) ${{m}_{{{H}_{2}}{{O}_{2}}}}=13.88$

Answer 1

Hint: Composition of solution can be described by expressing its concentration either qualitatively or quantitatively. There are several ways by which we can describe the concentration of the solution quantitatively. There are Mass percentage, volume percentage, mass by volume percentage, ppm, mole fraction, molarity, normality and molality.

Complete answer:
(1) Consider the equation,
${{H}_{2}}{{O}_{2}}\to {{H}_{2}}O+\dfrac{1}{2}{{O}_{2}}$
Stoichiometry of the above equation, 1 mole of ${{H}_{2}}{{O}_{2}}$ equal to $\dfrac{1}{2}mole$ of oxygen.
number of moles of oxygen = $\dfrac{28}{22.4}=1.25mole$ (from 28 labeled volume)
hence, number moles of hydrogen peroxide (${{H}_{2}}{{O}_{2}}$)= 2X number of moles of oxygen
= 2X1.25= 2.5 moles
The density of the solution = 26.5g/L
Then, The volume of the solution = 1L
Molarity of hydrogen peroxide, ${{M}_{{{H}_{2}}O}}_{_{2}}=\dfrac{moles\text{ }of\text{ }hydrogen\text{ }peroxide\text{ }}{volume\text{ }of\text{ }the\text{ }solution}=\dfrac{2.5mole}{1L}=2.5moles/L$
Hence, option A is correct.
(2) mass % of the component $%\dfrac{w}{v}=\dfrac{2.5molX34g/mol}{1000}X100=8.5$
(3) Given density of the solution = 26.5g/L
Then, Mass of solution = 26.5g
mass of water = mass of solution-moles of hydrogen peroxide molar mass of hydrogen peroxide
                    = 265g/LX1L-2.5molesX34g/mole=180g
Number of moles of water = \[\dfrac{mass\text{ }of\text{ }water}{mw\text{ }of\text{ }water}=\dfrac{180}{18}=10moles\]
Mole fraction of hydrogen peroxide ${{X}_{{{H}_{2}}{{O}_{2}}}}=\dfrac{moles\text{ }of\text{ }hydrogen\text{ }peroxide\text{ }}{Number\text{ }of\text{ }moles\text{ }of\text{ }water~+moles\text{ }of\text{ }hydrogen\text{ }peroxide}=\dfrac{2.5}{2.5+10}=0.2$
Hence, mole fraction of ${{H}_{2}}{{O}_{2}}$= 0.2
so, option C is also correct.
(4) molality of hydrogen peroxide, \[{{m}_{{{H}_{2}}{{O}_{2}}}}=\dfrac{moles\text{ }of\text{ }hydrogen\text{ }peroxide}{mass\text{ }of\text{ }water(Kg)}=\dfrac{2.5}{180Kg}X1000=13.88m\]

Hence, from the above calculations . options A,C and D correct.

Note:
Mole fraction unit is very useful in relating some physical properties of solution, say vapour pressure with concentration of the solutio and quite useful in describing the calculations involving gas mixtures.