How many grams of solid silver nitrate would you need to prepare 240.0mL of a 0.160M AgNO3 solution?
To calculate the number of grams of solid silver nitrate needed to prepare a certain volume and concentration of AgNO3 solution, you need to use the formula:
mass = Volume (in liters) × Concentration × Molar mass
Let's break down the process step-by-step:
1. Start by converting the volume from milliliters to liters:
Volume (in liters) = 240.0 mL × (1 L / 1000 mL)
Volume (in liters) = 0.240 L
2. Plug in the given values into the formula:
mass = 0.240 L × 0.160 M × Molar mass of AgNO3
3. Determine the molar mass of AgNO3.
AgNO3 consists of one silver atom (Ag) with a molar mass of 107.87 grams/mol, one nitrogen atom (N) with a molar mass of 14.01 grams/mol, and three oxygen atoms (O), each having a molar mass of 16.00 grams/mol.
Molar mass of AgNO3 = (1 × 107.87 g/mol) + (1 × 14.01 g/mol) + (3 × 16.00 g/mol)
Molar mass of AgNO3 = 169.87 g/mol
4. Substitute the values into the formula and calculate the mass of silver nitrate:
mass = 0.240 L × 0.160 M × 169.87 g/mol
mass = 6.167 g
Therefore, you would need approximately 6.167 grams of solid silver nitrate to prepare 240.0 mL of a 0.160 M AgNO3 solution.
M=mol/L
use stoichiometry
0.160M=0.160mol/L
find mol
0.160mol/L*0.240L=0.0384 mol
multiply by molar mass of silver nitrate
0.0384mol*169.87g/mol=6.523g AgNO3