Question

If \(10 . \mathrm{g}\) of \(\mathrm{AgNO}_{3}\) is available, what volume of \(0.25 \mathrm{M} \mathrm{AgNO}_{3}\) solution can be prepared?

Short answer

Using 10g of AgNO3, approximately 235.6 mL of a 0.25M AgNO3 solution can be prepared.

Step-by-step solution

Find the moles of AgNO3 in the given mass

To calculate the moles of AgNO3, we need to know its molar mass. The molar mass of AgNO3 is: \(M_{AgNO3} = M_{Ag} + M_{N} + 3 \times M_{O}\) Where: \(M_{Ag} \approx 107.87 g/mol\) \(M_{N} \approx 14.01 g/mol\) \(M_{O} \approx 16.00 g/mol\) Now, we can calculate the moles of AgNO3 using the given mass. moles = mass / molar mass

Calculate the molar mass of AgNO3

Using the molar masses of Ag, N, and O, we can find the molar mass of AgNO3: \(M_{AgNO3} = 107.87 + 14.01 + 3 \times 16.00 = 169.88 g/mol\)

Calculate the moles of AgNO3

Now, we can find the moles of AgNO3 using the given mass of 10g and the molar mass we calculated in Step 2. moles = mass / molar mass moles = 10g / 169.88 g/mol moles ≈ 0.0589 mol

Determine the volume of the 0.25M AgNO3 solution

We have the moles of AgNO3 and the desired molarity, so we can calculate the volume of the solution using the molarity formula: M = moles / volume Rearranging the equation to solve for the volume, we get: volume = moles / M Substituting the values we have for the moles and molarity: volume = 0.0589 mol / 0.25 M volume ≈ 0.2356 L

Convert the volume to milliliters (if necessary)

If we need the final volume in milliliters, we can convert it as follows: volume = 0.2356 L × 1000 mL/L volume ≈ 235.6 mL Therefore, using 10g of AgNO3, approximately 235.6 mL of a 0.25M AgNO3 solution can be prepared.