How many gm of `K_(2)Cr_(2)O_(7)` is present in 1 L of its N/10 solution in acid medium ?

Options

A. 4.9
B. 49
C. 0.49
D. 3.9

Answer

To find out how many grams of \( K_2Cr_2O_7 \) (potassium dichromate) are present in 1 L of its N/10 solution in acidic medium, we can follow these steps:

### Step 1: Understand the Concept of Normality
Normality (N) is defined as the number of equivalents of solute per liter of solution. An N/10 solution means that it is 0.1 N.

### Step 2: Determine the Valency Factor
In acidic medium, potassium dichromate \( K_2Cr_2O_7 \) reduces from chromium in the +6 oxidation state to +3. Each chromium atom undergoes a change of 3 in oxidation state. Since there are 2 chromium atoms in potassium dichromate, the total change in oxidation state (valency factor) is:
\[ \text{Valency Factor} = 2 \times 3 = 6 \]

### Step 3: Calculate the Equivalent Weight
The equivalent weight of a substance can be calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{Valency Factor}} \]
The molecular weight of \( K_2Cr_2O_7 \) is 294 g/mol. Therefore, the equivalent weight is:
\[ \text{Equivalent Weight} = \frac{294}{6} = 49 \text{ g/equiv} \]

### Step 4: Calculate the Number of Gram Equivalents in 1 L of 0.1 N Solution
Using the definition of normality:
\[ \text{Normality (N)} = \frac{\text{Number of Gram Equivalents}}{\text{Volume of Solution in Liters}} \]
For a 0.1 N solution:
\[ 0.1 = \frac{\text{Number of Gram Equivalents}}{1 \text{ L}} \]
Thus, the number of gram equivalents in 1 L of the solution is 0.1 equivalents.

### Step 5: Calculate the Weight of Potassium Dichromate
Using the relationship between weight, equivalent weight, and number of gram equivalents:
\[ \text{Weight} = \text{Number of Gram Equivalents} \times \text{Equivalent Weight} \]
Substituting the values:
\[ \text{Weight} = 0.1 \times 49 = 4.9 \text{ grams} \]

### Final Answer
The weight of \( K_2Cr_2O_7 \) present in 1 L of its N/10 solution in acidic medium is **4.9 grams**.

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