Calculate the number of grams of silver nitrate required to prepare 2.00L of 0.500 M AgNO3.
massAgNO3=formulamassAgNO3*Molarity*volume
Thank you so very much for your help.
To calculate the number of grams needed, we can use the formula:
grams = (Molarity) x (Volume in Liters) x (Molar Mass)
First, we need to find the molar mass of AgNO3:
AgNO3: (1 Ag) + (1 N) + (3 O) = 107.87 g/mol + 14.01 g/mol + (3 x 16.00 g/mol)
= 107.87 g/mol + 14.01 g/mol + 48.00 g/mol
= 169.88 g/mol
Now, we can plug the values into the formula:
grams = (0.500 M) x (2.00 L) x (169.88 g/mol)
grams = 0.500 x 2.00 x 169.88
grams = 169.88 g
Therefore, you would need 169.88 grams of silver nitrate to prepare 2.00L of 0.500 M AgNO3.
To calculate the number of grams of silver nitrate required to prepare the given solution, we need to use the formula:
Number of moles = concentration x volume
Step 1: Calculate the volume of the solution in liters.
The given volume is 2.00 L.
Step 2: Calculate the number of moles of silver nitrate.
Given concentration is 0.500 M (moles per liter).
Number of moles = concentration x volume
Number of moles = 0.500 M x 2.00 L = 1.00 moles
Step 3: Convert moles to grams.
We will use the molar mass of silver nitrate (AgNO3) to convert moles to grams.
The molar mass of AgNO3 is:
AgNO3 = (1 Ag atom x atomic mass of Ag) + (1 N atom x atomic mass of N) + (3 O atoms x atomic mass of O)
AgNO3 = (1 x 107.87 g/mol) + (1 x 14.01 g/mol) + (3 x 16.00 g/mol)
AgNO3 = 169.87 g/mol
Now, we can calculate the grams of silver nitrate required:
Number of grams = Number of moles x Molar mass
Number of grams = 1.00 moles x 169.87 g/mol
Number of grams = 169.87 grams
Therefore, to prepare 2.00 L of 0.500 M AgNO3 solution, you would need 169.87 grams of silver nitrate.