If 10. g AgNO3 is available, what volume of 0.25M AgNO3 solution can be prepared?

moles AgNO3 = grams/molar mass

M = moles/L.
You know M and moles, solve for L.

Thank You!

You're quite welcome.

235 ml

To determine the volume of 0.25M AgNO3 solution that can be prepared using 10g of AgNO3, we need to use the equation:

Molarity (M) = moles of solute / volume of solution (in liters)

First, we need to calculate the moles of AgNO3 using its molar mass. The molar mass of AgNO3 can be calculated as follows:

Molar mass (AgNO3) = (atomic mass of Ag * number of Ag atoms) + (atomic mass of N * number of N atoms) + (atomic mass of O * number of O atoms)

= (107.87 * 1) + (14.01 * 1) + (16.00 * 3)
= 107.87 + 14.01 + 48.00
= 169.88 g/mol

Next, we can calculate the moles of AgNO3 by dividing the given mass by the molar mass:

moles of AgNO3 = mass / molar mass
moles of AgNO3 = 10g / 169.88 g/mol

Now that we have the moles of AgNO3, we can calculate the volume of 0.25M AgNO3 solution using the equation mentioned earlier:

0.25M = moles of AgNO3 / volume of solution (in liters)

Rearranging the equation, we get:

Volume of solution = moles of AgNO3 / 0.25M

Substituting the calculated values, we get:

Volume of solution = (10g / 169.88 g/mol) / 0.25M

Simplifying further:

Volume of solution = (10 / 169.88) / 0.25
Volume of solution ≈ 0.0586 L

Therefore, approximately 0.0586 liters (or 58.6 mL) of 0.25M AgNO3 solution can be prepared using 10g of AgNO3.