Question

What is the mass of potassium dichromate crystals required to oxidize $750c{{m}^{3}}$ of $0.6M$ solution of Mohr’s salt if the molecular mass of potassium dichromate is 294 and molecular mass of Mohr’s salt is 392?
A. 0.39g
B. 0.37g
C. 22.05g
D. 2.2g

Answer 1

Hint: Consider the reaction of oxidation of Mohr’s salt by potassium dichromate. Determine the number of moles of potassium dichromate required to oxidize the given number of moles of Mohr’s salt.

Complete step by step answer:
First, we will find the number of moles of Mohr’s salt present before looking at the reaction that will decide how many moles of potassium dichromate will be required.
It is given that the volume that the solution of Mohr’s salt occupies is $750c{{m}^{3}}$ and the molarity of the solution is $0.6M$. Using the formula for molarity, we can easily find the number of moles of Mohr’s salt present.
\[\text{Molarity}=\dfrac{\text{no}\text{. of moles}}{\text{volume in liters}}\]
Converting the value for volume, we get:
\[\begin{align}
 & 0.6M=\frac{\text{no}\text{. of moles}}{0.75{{m}^{3}}} \\
 & \text{no}\text{.of moles = }0.6M\times 0.75{{m}^{3}} \\
 & \text{no}\text{.of moles = }0.455mol \\
\end{align}\]

Mohr’s salt is a double salt whose molecular formula is $FeS{{O}_{4}}\cdot {{(N{{H}_{4}})}_{2}}S{{O}_{4}}\cdot 6{{H}_{2}}O$. In this reaction the $F{{e}^{2+}}$ is oxidized to $F{{e}^{3+}}$ but the ammonium sulphate as well as the water of hydration (water molecules that hydrate the salt) remain unreacted. So, we will exclude them from the overall reaction. The overall reaction is:
\[{{K}_{2}}C{{r}_{2}}{{O}_{7}}+6FeS{{O}_{4}}+7{{H}_{2}}S{{O}_{4}}\to 3F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+{{K}_{2}}S{{O}_{4}}+C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+7{{H}_{2}}O\]

Let us focus only on the ferrous sulphate and the potassium dichromate molecules. We can see that one mole of potassium dichromate is enough to oxidize six moles of ferrous sulphate from Mohr’s salt. Thus, we can say that one mole of potassium dichromate can oxidize six moles of Mohr’s salt.

We have 0.455 moles of Mohr’s salt; so, let us calculate how many moles of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ will be required to oxidize the calculated amount of Mohr’s salt. We will find this value by cross multiplication. Consider that $x$ is the number of moles of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ that will be required to oxidize the given amount of salt
\[\begin{align}
  & \dfrac{\text{moles of }{{K}_{2}}C{{r}_{2}}{{O}_{7}}}{\text{moles of }FeS{{O}_{4}}}=\dfrac{\text{given moles of }{{K}_{2}}C{{r}_{2}}{{O}_{7}}}{\text{given moles of }FeS{{O}_{4}}} \\
 & \dfrac{1}{6}=\dfrac{x}{0.455} \\
 & x=0.075 \\
\end{align}\]

Now that we have the number of moles of potassium dichromate required, we can calculate the weight required using the formula to calculate the number of moles.
\[\text{no}\text{. of moles = }\dfrac{\text{given weight}}{\text{molecular weight}}\]

Here, the given weight will be the weight that we have to find out and we have been provided with the molecular weight of potassium dichromate as 294. Now, solving for the given weight.
\[\begin{align}
 & 0.075=\frac{\text{given weight}}{294} \\
 & \text{given weight = }0.075\times 294 \\
 & \text{given weight = }22.05g \\
\end{align}\]
So, the correct answer is “Option C”.

Additional Information:
We can also solve this sum by calculating the number of gram equivalents present. The ionic reaction of this reaction should be known to calculate the n-factor and then the number of moles can be found.

Note: The presence of sulphuric acid is essential for this reaction to take place, potassium dichromate only acts as a strong oxidizing agent in the presence of an acid. So, the acidic medium is indispensable. Also, do not get confused regarding the units of concentration. The concentration given here is in molarity.