What mass, in grams, g, of solid ammonium persulphate, (NH4)2S2O8 needed to prepare 250.0 mL of a 0.220 M solution?
Molar mass of (NH4)2S2O8 = 228.18 g/mol
How many mols do you need? That's L x M = 0.250 L x 0.220 M = ?
Then mols = g/molar mass. You have molar mass and mols, solve for grams.
0.250 L x 0.220 M = .055 mols. .055x228.18=12.55g?
If you typed in all of the significant figures n the problem, then the 0.220 M allows you to use 3 decimal places so I would round that to 12.6 g
To calculate the mass of solid ammonium persulphate needed, we can use the formula:
mass = volume x concentration x molar mass
The volume is given as 250.0 mL, the concentration as 0.220 M, and the molar mass as 228.18 g/mol.
Plugging these values into the formula, we get:
mass = 250.0 mL x 0.220 M x 228.18 g/mol
Now, let me do the math for you...
Hmm... I forgot to bring my calculator. Looks like I'll have to borrow one from Pennywise the Clown. Just kidding! Let's continue.
mass = 250.0 mL x 0.220 M x 228.18 g/mol
By multiplying these values together, we get:
mass = 1249.5 g
So, you would need approximately 1249.5 grams of solid ammonium persulphate to prepare 250.0 mL of a 0.220 M solution. Remember, though, that this is just an approximation. So, don't be too surprised if the real value differs slightly.
To calculate the mass of solid ammonium persulphate needed to prepare a desired volume and concentration of a solution, you can use the following formula:
Mass (g) = Volume (L) × Concentration (mol/L) × Molar mass (g/mol)
First, let's convert the volume from milliliters (mL) to liters (L):
250.0 mL ÷ 1000 = 0.250 L
Now we can substitute the given values into the formula:
Mass (g) = 0.250 L × 0.220 mol/L × 228.18 g/mol
Multiplying these values together gives us the mass in grams:
Mass (g) = 12.65 g
Therefore, you would need 12.65 grams of solid ammonium persulphate to prepare 250.0 mL of a 0.220 M solution.