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 amount of lithium nitrate (LiNO3) required to prepare a given volume and concentration of solution, you need to use the formula:
Mass = Volume × Concentration × Molar Mass
1. First, determine the molar mass of lithium nitrate (LiNO3):
- The atomic mass of lithium (Li) is approximately 6.94 g/mol.
- Nitrogen (N) has an atomic mass of approximately 14.01 g/mol.
- Oxygen (O) has an atomic mass of approximately 16.00 g/mol.
- Since there are three oxygen atoms in lithium nitrate, multiply its atomic mass (16.00 g/mol) by 3. Summing these values, we get:
Molar mass of LiNO3 = (6.94 g/mol) + (14.01 g/mol) + (16.00 g/mol × 3) = 85.94 g/mol
2. Next, substitute the given values into the formula:
Volume = 3.00 × 10^2 mL = 300 mL (convert mL to L by dividing by 1000)
3. The concentration is given as 2.00 × 10^–1 M.
4. Using the formula, the calculation will be:
Mass of LiNO3 = Volume × Concentration × Molar Mass
= 0.300 L × 2.00 × 10^–1 M × 85.94 g/mol
5. Substitute the values and calculate:
Mass of LiNO3 = 0.300 × 2.00 × 85.94 g
The answer is the calculated mass of lithium nitrate (LiNO3) in grams.
M = moles/L
solve for moles.
moles = grams/molar mass
solve for grams.