The mass of potassium dichromate crystal required to oxidise `750 cm^(3)` of `0.6` M Mohr's salt solution is `(molar mass=392)`

Options

A. (a)`0.49 g`
B. (b)`0.45 g`
C. (c )`22.05 g`
D. (d)`2.2 g`

Explanation

To solve the problem of finding the mass of potassium dichromate (K₂Cr₂O₇) required to oxidize 750 cm³ of 0.6 M Mohr's salt solution, we will follow these steps:

### Step 1: Calculate the number of moles of Mohr's salt.
1. **Convert the volume from cm³ to liters**:
\[ 750 \, \text{cm}^3 = 750 \, \text{mL} = \frac{750}{1000} \, \text{L} = 0.75 \, \text{L} \]

2. **Calculate the number of moles using molarity**:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (L)} = 0.6 \, \text{mol/L} \times 0.75 \, \text{L} = 0.45 \, \text{moles} \]

### Step 2: Write the balanced chemical equation.
The balanced reaction between potassium dichromate and Mohr's salt (FeSO₄) can be represented as:
\[ K_2Cr_2O_7 + 6FeSO_4 + 7H_2SO_4 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + 3Fe_2(SO_4)_3 + 7H_2O \]

### Step 3: Determine the stoichiometry of the reaction.
From the balanced equation, we see that:
- 1 mole of K₂Cr₂O₇ reacts with 6 moles of Mohr's salt.
- Therefore, for 1 mole of Mohr's salt, \(\frac{1}{6}\) moles of K₂Cr₂O₇ are required.

### Step 4: Calculate the moles of K₂Cr₂O₇ needed.
Using the number of moles of Mohr's salt calculated:
\[ \text{Moles of K}_2Cr_2O_7 = \frac{1}{6} \times 0.45 \, \text{moles} = 0.075 \, \text{moles} \]

### Step 5: Calculate the mass of K₂Cr₂O₇ required.
Using the formula:
\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]
Given that the molar mass of K₂Cr₂O₇ is 392 g/mol:
\[ \text{Mass of K}_2Cr_2O_7 = 0.075 \, \text{moles} \times 392 \, \text{g/mol} = 29.4 \, \text{grams} \]

### Final Answer:
The mass of potassium dichromate required to oxidize 750 cm³ of 0.6 M Mohr's salt solution is **29.4 grams**.

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