The weight of `AgNO_(3)` (Mol.wt = 170) present in 100 ml of 0.25 M solution in - grams

Options

A. `4.25`
B. 2
C. 17
D. `1.7`

Explanation

To find the weight of `AgNO3` in grams present in 100 ml of a 0.25 M solution, we can follow these steps:

### Step 1: Understand the definition of molarity
Molarity (M) is defined as the number of moles of solute per liter of solution.

### Step 2: Convert the volume of the solution from milliliters to liters
Given that the volume of the solution is 100 ml, we convert this to liters:
\[ \text{Volume in liters} = \frac{100 \, \text{ml}}{1000} = 0.1 \, \text{L} \]

### Step 3: Use the molarity to find the number of moles of `AgNO3`
The molarity of the solution is given as 0.25 M. We can use the formula for molarity to find the number of moles:
\[ \text{Molarity} = \frac{\text{Number of moles}}{\text{Volume in liters}} \]
Rearranging this gives us:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume in liters} \]
Substituting the known values:
\[ \text{Number of moles} = 0.25 \, \text{mol/L} \times 0.1 \, \text{L} = 0.025 \, \text{moles} \]

### Step 4: Calculate the weight of `AgNO3`
To find the weight of `AgNO3`, we use the formula:
\[ \text{Weight} = \text{Number of moles} \times \text{Molar mass} \]
Given that the molar mass of `AgNO3` is 170 g/mol, we can calculate:
\[ \text{Weight} = 0.025 \, \text{moles} \times 170 \, \text{g/mol} = 4.25 \, \text{grams} \]

### Final Answer
The weight of `AgNO3` present in 100 ml of a 0.25 M solution is **4.25 grams**.

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