You need to prepare 2 M solutions, with 10 g of solute. What volume of solution will you prepare for ammonium nitrate, NH4NO3?

How many moles do you have with 10 g NH4NO3? That will be moles = grams/molar mass.

M = moles/L soln. You know moles and M, solve for L. Note that this will be the total volume of the solution; i.e., not 10 g NH4NO3 + that much water.

To prepare a 2 M solution of ammonium nitrate (NH4NO3) with 10 g of solute, you'll need to follow the steps below:

Step 1: Determine the molar mass of ammonium nitrate (NH4NO3).
The molar mass of NH4NO3 is calculated as follows:
N (nitrogen) = 14.01 g/mol
H (hydrogen) = 1.01 g/mol (4 hydrogens in NH4)
O (oxygen) = 16.00 g/mol (3 oxygens in NO3)
Total molar mass = (14.01 x 1) + (1.01 x 4) + (16.00 x 3) = 80.04 g/mol

Step 2: Calculate the number of moles of ammonium nitrate (NH4NO3) required.
Number of moles = mass / molar mass
Number of moles = 10 g / 80.04 g/mol = 0.1249 mol

Step 3: Calculate the volume of solution required.
Molarity (M) = moles / volume (in liters)
Rearranging the formula, we have:
Volume (in liters) = moles / Molarity
Volume (in liters) = 0.1249 mol / 2 M = 0.06245 L

Step 4: Convert the volume from liters to milliliters (mL).
As there are 1000 mL in 1 L, the volume in mL is calculated as follows:
Volume (in mL) = Volume (in liters) x 1000 = 0.06245 L x 1000 = 62.45 mL

Therefore, to prepare a 2 M solution of ammonium nitrate (NH4NO3) with 10 g of solute, you will need to prepare 62.45 mL of solution.

To calculate the volume of a solution, you need to know the concentration of the solution (in this case, 2 M) and the amount of solute (in this case, 10 g).

Firstly, we need to determine the molar mass of ammonium nitrate (NH4NO3).

NH4NO3 consists of one nitrogen atom (N), two hydrogen atoms (H), and three oxygen atoms (O). The atomic masses of N, H, and O are approximately 14.01 g/mol, 1.01 g/mol, and 16.00 g/mol, respectively.

So, the molar mass of ammonium nitrate can be calculated as follows:

Molar mass of NH4NO3 = (1 * Molar mass of N) + (4 * Molar mass of H) + (3 * Molar mass of O)
= (1 * 14.01 g/mol) + (4 * 1.01 g/mol) + (3 * 16.00 g/mol)
= 80.04 g/mol

Now that we know the molar mass of ammonium nitrate is 80.04 g/mol, we can use this information to calculate the number of moles of solute (ammonium nitrate) in the 10 g.

Number of moles = Mass of solute / Molar mass
= 10 g / 80.04 g/mol
= 0.125 mol

Since you want to prepare a 2 M solution, which means there will be 2 moles of ammonium nitrate per liter of solution, we can use this information to calculate the volume of the solution.

Volume of solution (in liters) = Number of moles of solute / Concentration of solution
= 0.125 mol / 2 mol/L
= 0.0625 L

Therefore, to prepare a 2 M solution of ammonium nitrate with 10 g of solute, you would need to prepare 0.0625 liters (or 62.5 mL) of the solution.