how many grams of NaOH do you need to prepare a 3.00 x 10^2 mL stock solution of 2.00 M NaOH
mols NaOH you need = M x L = 2.00 x 0.300 = ?
Then grams = mols NaOH x M NaOH = ?
To determine the number of grams of NaOH needed to prepare a stock solution, you'll need to use the formula:
moles = concentration (M) × volume (L)
First, let's convert the given volume from mL to L:
3.00 x 10^2 mL = 3.00 x 10^2 ÷ 1000 = 0.300 L
Now we can calculate the moles of NaOH needed:
moles = 2.00 M × 0.300 L = 0.600 mol
Next, we need to convert moles to grams using the molar mass of NaOH:
molar mass of NaOH = (mass of Na) + (mass of O) + (mass of H)
= (23.0 g/mol) + (16.0 g/mol) + (1.0 g/mol)
= 40.0 g/mol
Finally, we can calculate the grams of NaOH needed:
grams = moles × molar mass of NaOH
= 0.600 mol × 40.0 g/mol
= 24.0 g
Therefore, you need 24.0 grams of NaOH to prepare a 3.00 x 10^2 mL stock solution of 2.00 M NaOH.
To calculate the number of grams of NaOH needed to prepare a stock solution, we need to use the formula:
grams = (volume × concentration × molar mass) / 1000
First, let's calculate the molar mass of NaOH:
- The atomic mass of Na (sodium) is approximately 23 grams/mol.
- The atomic mass of O (oxygen) is approximately 16 grams/mol.
- The atomic mass of H (hydrogen) is approximately 1 gram/mol.
Therefore, the molar mass of NaOH is:
molar mass = (23 grams/mol) + (16 grams/mol) + (1 gram/mol)
= 40 grams/mol
Now, let's substitute the given values into the formula:
grams = (300 mL × 2.00 mol/L × 40 grams/mol) / 1000
First, convert the volume from milliliters to liters:
volume = 300 mL / 1000 mL/L
= 0.300 L
Substitute the values into the formula:
grams = (0.300 L × 2.00 mol/L × 40 grams/mol) / 1000
= (0.300 × 2.00 × 40) grams
= 24 grams
Therefore, you would need 24 grams of NaOH to prepare a 300 mL stock solution of 2.00 M NaOH.