caluclate the volume in mL of 1.74 grams of sodium chloride from a 0.270 M solution
1.74g sodium chloride from a 0.270 M solution
7.10g ethonal from a 2.45 M solution
6.90g acetic acid (CH3COOH) from 0.606 M solution
For #1.
M = mols/L. You know M, you want to find L, that leaves moles that you don't know. How can you find that? That is mols NaCl = grams NaCl/molar mass NaCl. You know g and molar mass, solve for mols and substitute into the original M = mols/L. Solve for L and convert to mL.
To calculate the volume in mL of a certain amount of a substance from a given solution, you need to use the equation:
Volume (in mL) = (Amount of substance in grams) / (Molarity x Molecular weight)
1. For sodium chloride (NaCl):
- Amount of substance = 1.74 grams
- Molarity = 0.270 M
First, you need to find the molecular weight of sodium chloride:
- Sodium (Na) has a atomic weight of 22.99 g/mol
- Chloride (Cl) has an atomic weight of 35.45 g/mol
Therefore, the molecular weight of sodium chloride (NaCl) is 22.99 g/mol + 35.45 g/mol = 58.44 g/mol.
Now, you can calculate the volume:
Volume (in mL) = 1.74 g / (0.270 mol/L * 58.44 g/mol)
Volume (in mL) = 112.63 mL
So, the volume in mL of 1.74 grams of sodium chloride from a 0.270 M solution is approximately 112.63 mL.
2. For ethanol (C2H5OH):
- Amount of substance = 7.10 grams
- Molarity = 2.45 M
The molecular weight of ethanol (C2H5OH) is 46.07 g/mol.
Volume (in mL) = 7.10 g / (2.45 mol/L * 46.07 g/mol)
Volume (in mL) = 62.55 mL
So, the volume in mL of 7.10 grams of ethanol from a 2.45 M solution is approximately 62.55 mL.
3. For acetic acid (CH3COOH):
- Amount of substance = 6.90 grams
- Molarity = 0.606 M
The molecular weight of acetic acid (CH3COOH) is 60.05 g/mol.
Volume (in mL) = 6.90 g / (0.606 mol/L * 60.05 g/mol)
Volume (in mL) = 191.78 mL
So, the volume in mL of 6.90 grams of acetic acid (CH3COOH) from a 0.606 M solution is approximately 191.78 mL.