Question

Given,250ml of ${{H}_{2}}{{O}_{2}}$ solution produces 64 grams of ${{O}_{2}}$ . Molarity of ${{H}_{2}}{{O}_{2}}$ is-
(A) 14
(B) 15
(C) 8
(D) 16

Answer 1

Hint: The chemical reaction involved is:
     \[2{{H}_{2}}{{O}_{2(aq)}}\to 2{{H}_{2}}{{O}_{(l)}}+{{O}_{2(g)}}\]
Use the following formulae to solve this question-
-No. of moles$=\dfrac{\text{given mass}}{\text{molar mass}}$
-And molarity of a solution $=\dfrac{no.\text{ of moles of solute }}{volume\text{ of solution (in L)}}$

Complete answer:
Formation of ${{O}_{2}}$from ${{H}_{2}}{{O}_{2}}$ can be shown by the following equation-
     \[2{{H}_{2}}{{O}_{2(aq)}}\to 2{{H}_{2}}{{O}_{(l)}}+{{O}_{2(g)}}\]
Given mass of ${{O}_{2}}$gas produced is 64g.
Molar mass of ${{O}_{2}}$$=2\times 16=32g/mol$
Number of moles of ${{O}_{2}}$$=\dfrac{\text{given mass of }{{\text{O}}_{2}}}{\text{Molar mass of }{{\text{O}}_{2}}}=\dfrac{64g}{32g/mol}=2moles$
As 1 mole of ${{O}_{2}}$gas is produced from 2 moles of ${{H}_{2}}{{O}_{2}}$. Thus 2 moles of ${{O}_{2}}$gas are produced from 4moles of ${{H}_{2}}{{O}_{2}}$.
As we know that, molarity of a solution $=\dfrac{no.\text{ of moles of solute }}{volume\text{ of solution (in L)}}$
     \[\begin{align}
  & Molarity\text{ of }{{\text{H}}_{2}}{{O}_{2}}=\dfrac{\text{No}\text{. of moles of }{{\text{H}}_{2}}{{O}_{2}}}{Volume\text{ of }{{\text{H}}_{2}}{{O}_{2}}\text{ (in L)}} \\
 & Molarity\text{ of }{{\text{H}}_{2}}{{O}_{2}}=\dfrac{4\times 1000}{250} \\
 & Molarity\text{ of }{{\text{H}}_{2}}{{O}_{2}}=16M \\
\end{align}\]
Molarity of ${{H}_{2}}{{O}_{2}}$ comes out to be 16M.

So the correct answer is option (D) 16.

Note:
Coefficient written before the chemical entity in a chemical equation indicates the no. of moles of the substance in the chemical reaction used and produced.
 Molarity should not be confused with molality. Molality(m) is defined as the number of moles of a solute per kilogram of a solvent whereas Molarity(M) is defined as the number of moles of a solute added in a solution per liter of solution.
     \[molality(m)=\dfrac{{{n}_{solute}}}{{{w}_{solvent}}}\]
 Where ${{n}_{solute}}=$ number of moles of solute added
${{w}_{solvent}}=$ mass of solvent in kilograms
Therefore , the units of molality is mol/kg or $mol.k{{g}^{-1}}$.