What volume of 0.01 mol L-1 Li2CO3 solution would contain 0.035 mol lithium carbonate?
Give your answer as 2 sig fig (this is technically not correct), followed by a space then the units.
M = mols/L. You have mols and M, solve for L.
which one is M?
To find the volume of the Li2CO3 solution containing 0.035 mol, we can use the equation:
n = C × V
Where:
n is the amount of substance in moles,
C is the concentration in mol L^-1, and
V is the volume in liters.
Rearranging the equation, we get:
V = n / C
Plugging in the given values:
n = 0.035 mol
C = 0.01 mol L^-1
V = 0.035 mol / 0.01 mol L^-1
V = 3.5 L
However, since you requested the answer to be given with two significant figures, we need to round the answer:
V ≈ 3.5 L
So, the volume of the 0.01 mol L^-1 Li2CO3 solution containing 0.035 mol of lithium carbonate is approximately 3.5 L.