Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# 5.0 $cm^3$ of ${{H}_{2}}{{O}_{2}}$ liberates 0.508g of iodine from an acidified KI solution. The strength of ${{H}_{2}}{{O}_{2}}$ solution in terms of volume strength at STP is:A. 6.48 volumesB. 4.48 volumesC. 7.68 volumesD. None of these

Last updated date: 20th Jun 2024
Total views: 414.3k
Views today: 8.14k
Verified
414.3k+ views
Hint: To solve this question you should have basic knowledge of the term Normality, molarity, n-factor and volume strength. Volume strength of hydrogen peroxide is a term used to express concentration of ${{H}_{2}}{{O}_{2}}$ in terms of volumes of oxygen gas based on its decomposition to form water and oxygen.

Molarity is defined as the number of moles of solute dissolved in one liter of the solution. It is denoted by the letter M.
The reaction according to the given question is:
${{H}_{2}}{{O}_{2}}+2{{I}^{-}}+2{{H}^{+}}\to 2{{H}_{2}}O+{{I}_{2}}$
Molecular mass of iodine is 254${gm}/{mol.}\;$
Moles of iodine = $\dfrac{0.508}{254}moles$
Now, according to the concept, ${{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}$
Where, ${{M}_{1}}$ is Molarity of ${{H}_{2}}{{O}_{2}}$ solution.
Hence, ${{V}_{1}}$ is Volume of ${{H}_{2}}{{O}_{2}}$ solution
And ${{M}_{2}}$ is Molarity of iodine
And ${{V}_{2}}$ is 1000ml.
By putting values in the above equation,
${{M}_{1}}\times 5=\dfrac{0.508}{254}\times 1000$
Hence, ${{M}_{1}}$= 0.4M
Normality is defined as the number of gram or mole equivalents of solute present in one litre of a solution. It is denoted by the letter N. Equivalent weight is known as the mass of one equivalent, i.e. the mass of a substance taken under consideration which will combine with or displace a definite quantity of another substance in a chemical reaction. For an element the equivalent weight is nothing but the mass which combines with or displaces 1.008 gram of hydrogen or 8.0 gram of oxygen or 35.5 gram of chlorine. It is also known as gram equivalents.
To find the normality of the solution, $Normality=n-factor\times Molarity$
Where, N is the normality of the solution,
M is the molarity of the solution and n is the n-factor.
N-factor for acids, n-factor is defined as the number of ${{H}^{+}}$ ions replaced by 1 mole of acid in a reaction. For bases, it is defined as the number of ${{OH}^{-}}$ ions replaced by 1 mole of base in a reaction.
Normality of ${{H}_{2}}{{O}_{2}}$ is N.
n-factor for ${{H}_{2}}{{O}_{2}}$ is 2
Molarity of ${{H}_{2}}{{O}_{2}}$ is 0.4M
Hence, by putting values of all the unknown in this equation, $N=n\times M$,we can find N.
$N=2\times 0.4=0.8$
Hence, volume strength of ${{H}_{2}}{{O}_{2}}$ at STP is $=N\times 5.6$
Volume strength of ${{H}_{2}}{{O}_{2}}$ at STP is $=0.8\times 5.6$
= 4.48 volumes.

Hence, B is correct.

Note:
-Don’t get confused between molarity and molality. Molarity is defined as the number of moles of solute dissolved in one liter of the solution whereas molality is defined as the number of moles of the solute per kilogram of the solvent.
-N-factor is not equal to its basicity i.e. the number of moles of replaceable H+ atom present in one mole of acid.
-N-factor is not equal to its acidity i.e. the number of moles of replaceable OH- atoms present in one mole of base.