If you recovered 4.65 g of Cu from a 42.0 mL sample of material, what would the molarity be?
4.65 g Cu/ 63.5 g/mol = 0.0732 mol
0.0732 mol/ 42 mL x 1000 mL/ 1 L =1.74 M ~ 2 M Cu??? Is this correct sig fig because you go off of the 1 sig in 1,000 mL??? So you get 2 with rounding 1.74 with following sig fig rules..
I would say M = mols/L
mols - 0.0732
L - 0.0420
M - 1.74M and use all three digits because, by definition, 1000 mL = 1 L.
oh, but with 1 L it is still only one sig fig? right?
Yes, 1 L has only 1 s.f.; however, 42.0 mL = 0.0420 L and there are 3 s.f. in 4.65g and in 0.0420. The "1 L" is never used since the definition is M = mols/L. The "L" in the formula happens to be 0.0420.
Okay that makes sense but if you had 40 mL which is just .04 then it would just be 1 sig fig.
.04 L to be exact
As I undersand it, yes. 0.04 has 1 s.f.
To find the molarity, you need to divide the number of moles of solute (in this case, copper, Cu) by the volume of the solution in liters.
Given:
- Mass of Cu recovered = 4.65 g
- Molar mass of Cu = 63.5 g/mol
- Volume of the solution = 42.0 mL
First, calculate the number of moles of Cu:
4.65 g Cu / 63.5 g/mol = 0.0732 mol Cu
Next, convert the volume from milliliters to liters:
42.0 mL x (1 L / 1000 mL) = 0.0420 L
Finally, divide the number of moles by the volume in liters to get the molarity:
0.0732 mol / 0.0420 L = 1.74 M (rounding the answer to two significant figures)
Therefore, the molarity of the solution is approximately 1.74 M.