Calculate the mass of silver nitrate, AgNO3, FW = 169.9 g/mole, that must be used to prepare 500.0 mL of 0.250 M silver nitrate solution.
How many moles do you need? That is M x L = ?
Then moles = grams/molar mass. Solve for grams.
To calculate the mass of silver nitrate required, we can use the formula:
Mass = Volume × Concentration × Molar Mass
First, let's convert the volume from milliliters (mL) to liters (L):
500.0 mL = 500.0 mL ÷ 1000 = 0.500 L
Now, we can use the formula:
Mass = 0.500 L × 0.250 M × 169.9 g/mol
Mass = 21.2375 g
Therefore, the mass of silver nitrate required to prepare 500.0 mL of 0.250 M silver nitrate solution is approximately 21.24 g.
To calculate the mass of silver nitrate needed to prepare a solution, we can use the formula:
Mass = Volume x Concentration x Formula Weight
First, let's convert the given volume from milliliters (mL) to liters (L):
500.0 mL = 500.0 mL / 1000 = 0.5 L
Next, we can use the formula:
Mass = 0.5 L x 0.250 mol/L x 169.9 g/mol
The concentration of the solution is given as 0.250 M (moles per liter). The formula weight (FW) of silver nitrate is given as 169.9 g/mol.
Now, let's substitute the values into the formula:
Mass = 0.5 L x 0.250 mol/L x 169.9 g/mol
Mass = 21.2375 g
Therefore, you would need approximately 21.24 grams of silver nitrate to prepare 500.0 mL of 0.250 M silver nitrate solution.