Calculate the amount of the water which must be added to 2 ml of a solution of concentration of 40 mg silver nitrate per ml, yield a solution of concentration fo 16 mg silver nitrate per ml?

Answer

To solve the problem of calculating the amount of water that must be added to dilute a silver nitrate solution, we can follow these steps:

### Step 1: Understand the Initial Conditions
We have:
- Initial volume of the solution (V1) = 2 ml
- Initial concentration of silver nitrate (C1) = 40 mg/ml

### Step 2: Convert Concentration to Grams
Convert the concentration from mg/ml to grams/ml:
\[ C1 = 40 \, \text{mg/ml} = 40 \times 10^{-3} \, \text{g/ml} = 0.040 \, \text{g/ml} \]

### Step 3: Calculate the Initial Amount of Silver Nitrate
Calculate the total mass of silver nitrate in the initial solution:
\[ \text{Mass of AgNO}_3 = C1 \times V1 = 0.040 \, \text{g/ml} \times 2 \, \text{ml} = 0.080 \, \text{g} \]

### Step 4: Understand the Final Conditions
We want to achieve a final concentration (C2) of:
\[ C2 = 16 \, \text{mg/ml} = 16 \times 10^{-3} \, \text{g/ml} = 0.016 \, \text{g/ml} \]

### Step 5: Set Up the Equation for Final Volume
Let \( V \) be the volume of water added. The final volume (Vf) will be:
\[ Vf = V + V1 = V + 2 \, \text{ml} \]

### Step 6: Use the Final Concentration to Set Up the Equation
The total amount of silver nitrate remains the same, so we can set up the equation:
\[ \text{Mass of AgNO}_3 = C2 \times Vf \]
Substituting the known values:
\[ 0.080 \, \text{g} = 0.016 \, \text{g/ml} \times (V + 2) \]

### Step 7: Solve for V
Now, we can solve for \( V \):
\[ 0.080 = 0.016 \times (V + 2) \]
\[ 0.080 = 0.016V + 0.032 \]
Subtract 0.032 from both sides:
\[ 0.080 - 0.032 = 0.016V \]
\[ 0.048 = 0.016V \]
Now, divide both sides by 0.016:
\[ V = \frac{0.048}{0.016} = 3 \, \text{ml} \]

### Conclusion
The amount of water that must be added is **3 ml**.

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