How many grams of lithium nitrate, LiNO3, are required to prepare 3.00 × 102 mL of 2.00 × 10−1 M solution?
To calculate the number of grams of lithium nitrate (LiNO3) required, we need to use the formula:
Molarity (M) = moles of solute/volume of solution in liters
First, let's convert the given volume of solution from milliliters (mL) to liters:
3.00 × 10^2 mL = 3.00 × 10^2/1000 = 0.300 L
Now, let's rearrange the formula to calculate moles of solute:
moles of solute = Molarity × volume of solution in liters
moles of solute = 2.00 × 10^-1 mol/L × 0.300 L
moles of solute = 0.060 moles
Finally, we need to convert moles of solute to grams using the molar mass of lithium nitrate (LiNO3):
Molar mass of LiNO3 = (6.94 + 14.01 + 3 × 16.00) g/mol = 85.94 g/mol
grams of lithium nitrate = moles of solute × molar mass of lithium nitrate
grams of lithium nitrate = 0.060 moles × 85.94 g/mol
grams of lithium nitrate = 5.16 grams
Therefore, 5.16 grams of lithium nitrate are required to prepare 3.00 × 10^2 mL of a 2.00 × 10^-1 M solution.
To calculate the grams of lithium nitrate required to prepare a given solution, you will need to use the formula:
Mass (g) = Volume (L) × Concentration (M) × Molar Mass (g/mol)
1. Determine the molar mass of lithium nitrate (LiNO3):
- Lithium (Li) has a molar mass of approximately 6.94 g/mol.
- Nitrogen (N) has a molar mass of approximately 14.01 g/mol.
- Oxygen (O) has a molar mass of approximately 16.00 g/mol.
- Therefore, the molar mass of LiNO3 is:
Molar Mass (LiNO3) = (6.94 g/mol) + (14.01 g/mol) + (3 × 16.00 g/mol) = 85.94 g/mol
2. Convert the volume of the solution from milliliters (mL) to liters (L):
3.00 × 102 mL = 3.00 × 10-1 L
3. Plug the values into the formula:
Mass (g) = (3.00 × 10-1 L) × (2.00 × 10-1 M) × (85.94 g/mol)
4. Calculate the mass of lithium nitrate:
Mass (g) = 6.4752 g
Therefore, approximately 6.48 grams of lithium nitrate (LiNO3) are required to prepare 3.00 × 102 mL of a 2.00 × 10-1 M solution.
How many moles do you need in the final solution? That is M x L = 0.2M x 0.03L = ?
Then moles = grams/molar mass