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When calculating the molar mass of a compound like \(\text{AgNO}_3\), you need to add up the atomic masses of all the elements present. For \(\text{AgNO}_3\): Silver (Ag) has an atomic mass of 107.87 g/mol, Nitrogen (N) has an atomic mass of 14.01 g/mol, and Oxygen (O) has an atomic mass of 16.00 g/mol. Since there are three oxygen atoms in \(\text{AgNO}_3\), you multiply the atomic mass by 3. The molar mass of \(\text{AgNO}_3\) is calculated as follows: \[ \text{Molar mass} = 107.87 (\text{Ag}) + 14.01 (\text{N}) + 3 \times 16.00 (\text{O}) = 169.87 \text{ g/mol} \] Understanding the molar mass is crucial, as it serves as a bridge between the mass of a substance and the amount in moles.

To convert a given mass into moles, you use the molar mass calculated previously. The formula for this conversion is: \[ \text{Moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \] In the exercise, we need to convert 1.50 grams of \(\text{AgNO}_3\) into moles. Using the molar mass of 169.87 g/mol: \[ \text{Moles of AgNO}_3 = \frac{1.50 \text{ g}}{169.87 \text{ g/mol}} = 0.00883 \text{ mol} \] This step is fundamental in solution concentration problems, as it allows you to determine the number of moles of a solute present in a given mass.

Before calculating concentration, it is often necessary to convert volumes into the proper units. Molarity, for example, requires volume in liters. To convert milliliters (mL) to liters (L), use the conversion factor 1000 mL = 1 L. The formula is: \[ \text{Volume (L)} = \frac{\text{Volume (mL)}}{1000} \] In this exercise, we convert 25.0 mL of solution to liters: \[ 25.0 \text{ mL} = \frac{25.0}{1000} \text{ L} = 0.0250 \text{ L} \] Accurate conversion of volume is essential for determining the molarity of a solution.

Molarity (M) is a key concept in chemistry. It measures the concentration of a solution, defined as the number of moles of solute per liter of solution. The formula for calculating molarity is: \[ \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \] In the final step of the exercise, we calculate the molarity of the \(\text{AgNO}_3\) solution: \[ \text{Molarity} = \frac{0.00883 \text{ mol}}{0.0250 \text{ L}} = 0.3532 \text{ M} \] Understanding molarity helps standardize solutions for various lab and practical applications.