Calculate the molarity of the K+ ion in a solution prepared by adding of 5.55 g of K2SO4 to enough water to form 0.500 L of solution.
mols K2SO4 = grams/molar mass
M K2SO4 = mols/L
M K^+ = twice that because there are 2 K ions in 1 molecule of K2SO4.
25 grams of K2SO4 is added to 2500 mL of water, what will be the molarity of this solution?
To calculate the molarity of the K+ ion in the solution, you'll need to follow the steps below:
Step 1: Determine the number of moles of K2SO4.
To do this, you need to divide the mass of K2SO4 by its molar mass.
The molar mass of K2SO4 can be calculated as follows:
2(K) + 1(S) + 4(O)
2(39.10 g/mol) + 32.06 g/mol + 4(16.00 g/mol)
78.20 g/mol + 32.06 g/mol + 64.00 g/mol
174.26 g/mol
Now, divide the mass of K2SO4 (5.55 g) by its molar mass (174.26 g/mol) to find the number of moles:
5.55 g / 174.26 g/mol = 0.0319 mol
Step 2: Determine the number of moles of K+ ions.
Since there are two K+ ions in each K2SO4 molecule, multiply the number of moles of K2SO4 by 2:
0.0319 mol * 2 = 0.0638 mol
Step 3: Determine the volume of the solution in liters.
The volume of the solution is given as 0.500 L.
Step 4: Calculate the molarity.
Molarity (M) is calculated as moles of solute divided by liters of solution.
Molarity = moles of K+ ions / volume of solution in liters
Molarity = 0.0638 mol / 0.500 L = 0.1276 M
Therefore, the molarity of the K+ ion in the solution is 0.1276 M.