What mass of silver nitrate (AgNO3) is required to prepare 400 g of a 3.50% solution of

AgNO3

3.50% AgNO3 means 3.50g AgNO3/100 g soln. You want 400 g; therefore, you will need

3.50g x (400/100) = ?? g AgNO3.

For your second post, this link will work all of your stoichiometry problems. Remember M x L = moles.
http://www.jiskha.com/science/chemistry/stoichiometry.html

How many milliliters of a 0.9 molar HCl

solution are needed to react completely with
15 grams of zinc according to the following
equation?
? HCl + ? Zn → ? ZnCl2 + ? H2
Answer in units of m

To determine the mass of silver nitrate (AgNO3) required to prepare a 3.50% solution, we need to use the equation:

Mass of solute = volume of solution × concentration

Here, the volume of solution is not given, but we do have the mass of the solution, which is 400 g. To proceed with the calculation, we need to assume a specific volume, such as 100 mL (which is often used as a standard in chemistry).

1. Calculate the volume of the solution using the equation:

Volume = Mass / Density

Since the density of a 3.50% AgNO3 solution is not provided, let's assume it to be 1.00 g/mL for simplicity. Therefore:

Volume = Mass / Density = 400 g / 1.00 g/mL = 400 mL

2. Once we have the volume of the solution, we can calculate the mass of AgNO3 required using the equation:

Mass of AgNO3 = volume × concentration

Mass of AgNO3 = 400 mL × 0.035 (since 3.50% concentration is equivalent to 0.035 in decimal form)

Mass of AgNO3 = 14 g

Therefore, to prepare a 400 g solution of 3.50% AgNO3, you would need 14 grams of silver nitrate (AgNO3).

To calculate the mass of silver nitrate required to prepare a given solution, we need to use the formula:

Mass of solute = (Percent concentration / 100) * Mass of solution

In this case, the mass of the solution is given as 400 g, and the percent concentration is given as 3.50%.

Step 1: Convert the percent concentration to a decimal.

3.50% = 3.50 / 100 = 0.035

Step 2: Use the formula to calculate the mass of the solute (silver nitrate).

Mass of solute = 0.035 * 400 g = 14 g

So, you would require 14 grams of silver nitrate (AgNO3) to prepare 400 g of a 3.50% solution of AgNO3.