What is the molarity of a 5.00 x 102 ml solution containing 2490 g of KI?
how many moles of KI?
M = moles/liter
I'm dumb. I'm sorry I still don't understand.
5.00*10^2ml = 500ml = .5L
KI has molar mass 166, so you have 15 moles of KI
15moles/.5L = 30moles/L = 30M
cinchy, right?
How'd you get 15 moles?
To calculate the molarity of a solution, we need to know the amount of solute (in moles) and the volume of the solution (in liters).
Step 1: Convert the given volume from milliliters (ml) to liters (L).
5.00 x 10^2 ml = 5.00 x 10^2 / 1000 = 0.500 L
Step 2: Calculate the amount of solute (in moles) using the given mass and molar mass.
First, we need to convert the given mass of KI to moles. The molar mass of KI can be determined by adding the atomic masses of potassium (K) and iodine (I).
The atomic mass of K is 39.10 g/mol, and the atomic mass of I is 126.90 g/mol.
Molar mass of KI = (1 x 39.10 g/mol) + (1 x 126.90 g/mol) = 39.10 g/mol + 126.90 g/mol = 166.00 g/mol
Now, we can calculate the amount of KI in moles:
Amount of KI (in moles) = mass of KI / molar mass of KI
Amount of KI (in moles) = 2490 g / 166.00 g/mol
Amount of KI (in moles) ≈ 15.00 mol
Step 3: Calculate the molarity using the formula:
Molarity (M) = amount of solute (in moles) / volume of solution (in liters)
Molarity (M) = 15.00 mol / 0.500 L
Molarity (M) = 30.0 M
So, the molarity of the 5.00 x 10^2 ml solution containing 2490 g of KI is approximately 30.0 M.