How many grams of KF do you need to weigh in order to prepare 500. mL of solution with a concentration of 0.250 M?
M = mols/L and rearrange to mols = MxL
and mols = grams/molar mass
Therefore, since mols = mols we can set those equal to each other to obtain
mols = M x L = mols = grams/molar mass and
L x M x molar mass = grams.
Plug in the numbers and solve for grams.
I'm sorry, I don't even know where I should start plugging in. I got lost when you rearranged to mols = MxL.
M=mol/L
L*M = mols
just multiply both sides by L
Just use what I ended up with; i.e.,
M x L x molar mass = grams.
You have M, L, and molar mass. Solve for grams.
Okay...
M x L x mm = grams
0.250 M x 0.5 L x 57.9g = 7.24 grams
math looks fine but as for significant figures I'm not so sure.
Is that it for the problem, or is there another step I have to do?
That is the answer for the problem but as for significant figures it just means how many numbers you show for the answer.
for example if your answer is 8.93383494 right ? but the initial numbers given is 33.28 and 8.9 and because the 8.9 has the least significant figures (2) vs the 33.28's (4), then the answer would be 8.9.
So 7.2 g would be correct then?
To find out how many grams of KF you need for the solution, you need to know the molar mass of KF and use the equation:
Molarity (M) = moles of solute / liters of solution
If you rearrange the equation, you can solve for moles of solute:
moles of solute = Molarity (M) x liters of solution
First, convert the milliliters of solution to liters:
500 mL = 500 mL / 1000 mL/L = 0.5 L
Now you can calculate the moles of solute:
moles of solute = 0.250 M x 0.5 L = 0.125 moles
To convert moles of solute to grams, you need to know the molar mass of KF, which is made up of potassium (K) and fluorine (F). The atomic masses of K and F are 39.10 g/mol and 18.99 g/mol, respectively.
Molar mass of KF = (39.10 g/mol) + (18.99 g/mol) = 58.09 g/mol
Finally, you can calculate the grams of KF needed:
grams of KF = moles of solute x molar mass of KF
= 0.125 moles x 58.09 g/mol
= 7.26 grams
Therefore, you would need to weigh 7.26 grams of KF to prepare 500 mL of a 0.250 M solution.