determine the molarity of a solution of silver nitrate prepared by dissolving 5g of silver nitrate in 25.0mL of water
mols AgNO3 = grams/molar mass
Then M = mols/L solution.
Technically this problem can't be worked because the definition is mols/L SOLUTION and when you put 5.0 g AgNO3 in 25.0 mL water that DOESN'T make 25.0 mL solution but the solution volume is slight more than 25.0 mL. When someone makes up a problem like this, if they quote it as above then the density of the solution should be stated and you get the volume of the solution from the density of the solution and the mass of the solution.
To determine the molarity of a solution, you need to know the number of moles of solute (in this case, silver nitrate) and the volume of the solution.
First, let's determine the number of moles of silver nitrate.
The molar mass of silver nitrate (AgNO3) can be calculated by summing the atomic masses of its individual elements:
Ag (silver) = 107.87 g/mol
N (nitrogen) = 14.01 g/mol
O (oxygen) = 16.00 g/mol
Molar mass of AgNO3 = (1 x Ag) + (1 x N) + (3 x O) = 107.87 + 14.01 + (3 x 16.00) = 169.87 g/mol
Given that you have 5g of silver nitrate, you can calculate the number of moles using its molar mass:
Number of moles = Mass (g) / Molar mass (g/mol) = 5g / 169.87 g/mol
Next, let's determine the volume of the solution in liters. The volume is given as 25.0 mL, so we need to convert it to liters:
Volume (L) = 25.0 mL / 1000 mL/L = 0.025 L
Finally, we can calculate the molarity by dividing the number of moles of silver nitrate by the volume of the solution:
Molarity (M) = Number of moles / Volume (L)
Now, let's plug in the values:
Molarity = (5g / 169.87 g/mol) / 0.025 L
After performing the calculations, you'll find the molarity of the silver nitrate solution.