The volume of 0.1M `AgNO_(3)` which is required by 10 ml of 0.09 M `K_(2)CrO_(4)` to precipitate all the chromate as `Ag_(2)CrO_(4)` is

Options

A. 9 ml
B. 18 ml
C. 20 ml
D. 36 ml

Answer

To solve the problem, we need to determine the volume of 0.1 M AgNO₃ required to completely precipitate all the chromate ions from a 10 ml solution of 0.09 M K₂CrO₄ as Ag₂CrO₄.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction between AgNO₃ and K₂CrO₄ can be represented as:
\[ 2 \text{AgNO}_3 + \text{K}_2\text{CrO}_4 \rightarrow \text{Ag}_2\text{CrO}_4 \downarrow + 2 \text{KNO}_3 \]
From the balanced equation, we see that 2 moles of AgNO₃ react with 1 mole of K₂CrO₄ to form 1 mole of Ag₂CrO₄.

2. **Calculate the Number of Moles of K₂CrO₄**:
The molarity (M) of K₂CrO₄ is 0.09 M, and the volume (V) is 10 ml (which is 0.01 L). The number of moles (n) can be calculated using the formula:
\[ n = M \times V \]
\[ n = 0.09 \, \text{mol/L} \times 0.01 \, \text{L} = 0.0009 \, \text{mol} \]

3. **Determine the Moles of AgNO₃ Required**:
From the balanced equation, 2 moles of AgNO₃ are required for every 1 mole of K₂CrO₄. Therefore, the moles of AgNO₃ required will be:
\[ \text{Moles of AgNO}_3 = 2 \times \text{Moles of K}_2\text{CrO}_4 = 2 \times 0.0009 \, \text{mol} = 0.0018 \, \text{mol} \]

4. **Calculate the Volume of AgNO₃ Solution Required**:
We know the molarity of AgNO₃ is 0.1 M. We can use the formula:
\[ \text{Volume (L)} = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.0018 \, \text{mol}}{0.1 \, \text{mol/L}} = 0.018 \, \text{L} \]
To convert this to milliliters:
\[ \text{Volume (ml)} = 0.018 \, \text{L} \times 1000 \, \text{ml/L} = 18 \, \text{ml} \]

### Final Answer:
The volume of 0.1 M AgNO₃ required to precipitate all the chromate as Ag₂CrO₄ is **18 ml**.