Calculate the mass of silver nitrate needed to prepare a solution of concentration 0.01 moles per litre
I can't answer this without knowing how much of the solution you wish to prepare.
If you want it to be 0.01M and you want, say, 500 mL, then
moles needed = M x L = 0.01M x 0.5L = 0.005 moles.
Then moles = grams/molar mass and solve for grams.
Total silver nitrate volume =500 mL
To calculate the mass of silver nitrate needed to prepare a solution of concentration 0.01 moles per liter, we need to use the equation:
mass = moles * molar mass
First, let's start by finding the molar mass of silver nitrate (AgNO3). The molar mass is the sum of the atomic masses of each element in the compound.
Ag (silver) has an atomic mass of 107.87 g/mol
N (nitrogen) has an atomic mass of 14.01 g/mol
O (oxygen) has an atomic mass of 16.00 g/mol
So, the molar mass of AgNO3 is:
(107.87 g/mol) + (14.01 g/mol) + (3 * 16.00 g/mol) = 169.87 g/mol
Now, we can calculate the mass of silver nitrate needed using the equation mentioned earlier:
mass = moles * molar mass
Here, we have the concentration given in moles per liter. So, we can express it as moles = concentration * volume (in liters).
In this case, the concentration is 0.01 moles per liter. Let's assume we want to prepare a solution with a volume of 1 liter.
moles = 0.01 moles/L * 1 L = 0.01 moles
Now, we can substitute the values into the equation:
mass = 0.01 moles * 169.87 g/mol = 1.6987 grams
Therefore, you would need approximately 1.6987 grams of silver nitrate to prepare a solution with a concentration of 0.01 moles per liter.