How much `K_(2)Cr_(2)O_(7) (M.W. = 294.19)` is required to prepare one litre of `0.1N` solution?

Options

A. `9.8063g`
B. `7.3548g`
C. `3.6774g`
D. `4.903g`

Explanation

To determine how much \( K_2Cr_2O_7 \) (potassium dichromate) is required to prepare one liter of a \( 0.1N \) solution, we can follow these steps:

### Step 1: Understand Normality
Normality (N) is defined as the number of gram equivalents of solute per liter of solution. In this case, we need to find out how many gram equivalents of \( K_2Cr_2O_7 \) are present in 1 liter of a \( 0.1N \) solution.

### Step 2: Set Up the Equation
Given:
- Normality (N) = 0.1
- Volume of solution (L) = 1

Using the formula for normality:
\[ N = \frac{\text{Number of gram equivalents}}{\text{Volume of solution in liters}} \]
We can rearrange this to find the number of gram equivalents:
\[ \text{Number of gram equivalents} = N \times \text{Volume} = 0.1 \times 1 = 0.1 \text{ equivalents} \]

### Step 3: Calculate the Equivalent Weight
To find the weight of \( K_2Cr_2O_7 \), we need to know its equivalent weight. The equivalent weight can be calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n}} \]
where \( n \) is the number of electrons exchanged in the reaction.

For \( K_2Cr_2O_7 \), when it reacts with \( H^+ \) ions, it reduces from \( Cr^{6+} \) to \( Cr^{3+} \), which involves the transfer of 6 electrons. Therefore, \( n = 6 \).

The molar mass of \( K_2Cr_2O_7 \) is given as \( 294.19 \, g/mol \).

Now we can calculate the equivalent weight:
\[ \text{Equivalent weight} = \frac{294.19}{6} \approx 49.03 \, g/equiv \]

### Step 4: Calculate the Weight of \( K_2Cr_2O_7 \)
Now we can find the weight of \( K_2Cr_2O_7 \) required for 0.1 equivalents:
\[ \text{Weight} = \text{Number of equivalents} \times \text{Equivalent weight} \]
\[ \text{Weight} = 0.1 \times 49.03 \approx 4.903 \, g \]

### Step 5: Conclusion
Thus, the weight of \( K_2Cr_2O_7 \) required to prepare one liter of a \( 0.1N \) solution is approximately \( 4.9 \, g \).