Calculate the mass of KH2PO4 (136.09 g/mole) you will need to prepare 450 mL of the 0.05 M solution.
C=m/V
0.05=m/.450L
0.05*.450=m
0.0225=m
Is 0.0225g correct ?
No, 0.0225g is not the correct answer. To calculate the mass needed to prepare the solution, you need to use the formula:
mass = moles x molar mass
First, convert the volume from mL to L:
450 mL = 450/1000 = 0.45 L
Then, use the formula C = moles/L to find the moles of KH2PO4 needed:
0.05 M = moles/0.45 L
Solving for moles gives:
moles = 0.05 M x 0.45 L = 0.0225 moles
Finally, multiply the moles by the molar mass of KH2PO4 to find the mass needed:
mass = 0.0225 moles x 136.09 g/mol ≈ 3.072 g
Therefore, the correct mass of KH2PO4 needed to prepare the 0.05 M solution in 450 mL is approximately 3.072 grams.
No, 0.0225 g is not the correct answer. To calculate the mass of KH2PO4 needed, we can use the formula:
Mass = molarity x volume x molar mass
Given:
Molarity (M) = 0.05 M
Volume (V) = 450 mL = 0.45 L
Molar mass (MM) = 136.09 g/mol
Now, let's substitute these values into the formula:
Mass = 0.05 M x 0.45 L x 136.09 g/mol
Calculating this:
Mass = 3.06 g
Therefore, the mass of KH2PO4 needed to prepare 450 mL of the 0.05 M solution is 3.06 grams.