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25 ml of 0.08 N Mohr’s salt solution is oxidised by 20 ml of ${{\text{K}}_{2}}\text{C}{{\text{r}}_{2}}{{\text{O}}_{7}}$ in acidic medium. The mass of Mohr’s salt present in 500 cc is:
A. 3.96 g
B. 15.7 g
C. 39.6 g
D. 39.2 g

Last updated date: 17th Jun 2024
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Hint: The question deals with volumetric analysis and equivalents of the ions and compounds. We have to find the equivalents of Mohr’s salt and then its equivalent mas of it to obtain the mass of it present in 500 cc or 500 mL. (where 1 cc is equal to 1 mL).

Complete answer:
The acidic solution of potassium dichromate ${{\text{K}}_{2}}\text{C}{{\text{r}}_{2}}{{\text{O}}_{7}}$ is a strong oxidising agent and is reduced by ferrous ions to a green chromic salt when added to ferrous ammonium sulphate or Mohr's salt ${{\left( \text{N}{{\text{H}}_{4}} \right)}_{2}}\text{Fe}{{\left( \text{S}{{\text{O}}_{4}} \right)}_{2}}.6{{\text{H}}_{2}}\text{O}$ solution having diluted ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$. The reaction will be $\text{6F}{{\text{e}}^{2+}}+\text{C}{{\text{r}}_{2}}\text{O}_{7}^{2-}+14{{\text{H}}^{+}}\to 6\text{F}{{\text{e}}^{+3}}+2\text{C}{{\text{r}}^{+3}}+7{{\mathrm H }_{2}}{\mathrm O}$. By the involved reaction, we will find the equivalent mass of Mohr’ salt. The n-factor of Mohr’ salt is 1 as the change in oxidation state of Mohr’s salt is 1 from +2 to +3. ($\text{F}{{\text{e}}^{+2}}\to \text{F}{{\text{e}}^{+3}}+{{\text{e}}^{-}}$)

The molecular mass of Mohr’s salt is
- Atomic mass of $\text{Fe}$ is 56 grams.
- Atomic mass of $\text{H}$ is 1 gram.
- Atomic mass of $\text{N}$ is 14 grams.
- Atomic mass of $\text{S}$ is 32 grams.
- Atomic mass of$\text{O}$ is 16 grams.
Mass of Mohr’s salt is $\left[ 2\times \left( 14+4 \right)+56+2\times \left( 32+4\times 16 \right)+6\left( 2+16 \right) \right]$ will be 392 grams. The equivalent mass of Mohr’s salt is also 392 grams.
Now, let us find the equivalents of Mohr’ salt: 25ml of 0.08N Mohr’s salt solution corresponds to $25\times 0.08\times {{10}^{-3}}$$(\text{number of equivalents = Volume}\times \text{ Normality}\times {{10}^{-3}})$ be equal to 0.002 equivalents.
Mass of Mohr’s salt is $\left( \text{molar mass }\times \text{ equivalents} \right)$ will be $\left( 392\times 0.002 \right)$ is equal to 0.784 grams. 25 mL solution contains 0.784 g of Mohr's salt, then 500 mL of solution will contain $\left( \dfrac{0.784}{25} \right)\times 500$ is equal to 15.7 g of Mohr's salt.
The mass of Mohr’s salt present in 500 cc is 15.7 grams
So, the correct answer is “Option B”.

Note: The volume of potassium dichromate is given as additional information. The volume of potassium dichromate will be needed when we need to find the normality of ${{\text{K}}_{2}}\text{C}{{\text{r}}_{2}}{{\text{O}}_{7}}$ in the solution.