what is the molarity of iodine ions in a 100 mL solution containing 2.50g of MgI2?
2.5g X 1mol MgI2/ molar mass = mol MgI2
Molarity= mol of solute/ volume of solution (in L)
Phillip has calculated the molarity of MgI2 correctly. The problem asks for molarity of iodide ions and that will be just twice the molarity of MgI2.
To determine the molarity of iodine ions in the solution, we need to follow a series of steps:
Step 1: Calculate the number of moles of MgI2.
The molar mass of MgI2 is the sum of the atomic masses of magnesium (Mg) and iodine (I) multiplied by two since there are two iodine atoms in MgI2.
Molar mass of MgI2 = (24.31 g/mol + 2 * 126.90 g/mol) = 278.11 g/mol
To calculate the number of moles, divide the mass of MgI2 by its molar mass.
Number of moles of MgI2 = 2.50 g / 278.11 g/mol
Step 2: Convert the volume of the solution to liters.
The given volume is 100 mL. Since molarity is expressed in moles per liter (mol/L), we need to convert mL to L.
1 L = 1000 mL
Volume of the solution = 100 mL / 1000 mL/L = 0.100 L
Step 3: Calculate the molarity of iodine ions.
The molarity (M) is defined as moles of solute divided by the volume of the solution in liters.
Molarity of iodine ions = (Number of moles of MgI2 x 2) / Volume of the solution
Substitute the values into the equation:
Molarity of iodine ions = (2.50 g / 278.11 g/mol) x 2 / 0.100 L
Calculate the value to find the molarity of iodine ions in the solution.