3. Find the molarity of a solution made from 734 g of Li2SO4 and 2.5 L of water
mol wt of Li2SO4 = 109.9g/mole
So, you have 734/109.9 = 6.68 moles
6.68moles/2.5L = 2.67M solution
assuming the addition of the solute does not materially affect the volume of the solution.
To find the molarity of a solution, we need to first calculate the number of moles of the solute (Li2SO4) and then divide by the volume of the solution (in liters).
1. Calculate the number of moles of Li2SO4:
First, we need to determine the molar mass of Li2SO4.
The atomic masses of lithium (Li), sulfur (S), and oxygen (O) are:
Li = 6.94 g/mol
S = 32.07 g/mol
O = 16.00 g/mol
Since there are 2 lithium atoms, 1 sulfur atom, and 4 oxygen atoms in Li2SO4, we can calculate the molar mass:
Molar mass of Li2SO4 = (2 * Li) + S + (4 * O)
= (2 * 6.94 g/mol) + 32.07 g/mol + (4 * 16.00 g/mol)
= 109.94 g/mol
Now, calculate the number of moles of Li2SO4:
Number of moles = Mass (g) / Molar mass (g/mol)
= 734 g / 109.94 g/mol
≈ 6.675 moles
2. Calculate the volume of the solution in liters:
The volume of the solution is given as 2.5 L.
3. Calculate the molarity of the solution:
Molarity (M) = Number of moles / Volume (L)
= 6.675 moles / 2.5 L
≈ 2.67 M
Therefore, the molarity of the solution made from 734 g of Li2SO4 and 2.5 L of water is approximately 2.67 M.