How many grams of potassium iodide, KI, are required to make 2.0 L of a 0.1M solution?
Molarity = moles/ Liters
You are given the Molarity (0.1M) and Liters (2.0L). So, plug those into the formula and solve for moles.
0.1M = moles/ 2.0L
moles = 0.2mol
The molecular weight of KI is 166.0028. Convert moles to grams.
0.2mol* (166.0028g/1mol) = 3.2g KI
To calculate the number of grams of potassium iodide (KI) required to make a 0.1M solution in 2.0 L, you will need to use the formula:
Molarity (M) = moles of solute / volume of solution (L)
First, let's rearrange the formula to solve for moles of solute:
moles of solute = Molarity (M) x volume of solution (L)
Next, we can calculate the moles of KI:
moles of KI = 0.1 M x 2.0 L = 0.2 moles
Finally, we can convert moles of KI to grams using its molar mass. The molar mass of KI is approximately 166 grams/mol:
grams of KI = moles of KI x molar mass of KI = 0.2 moles x 166 grams/mol
Therefore, you will need approximately 33.2 grams of potassium iodide to make a 0.1M solution in 2.0 L.
To calculate the number of grams of potassium iodide (KI) required to make a 0.1M solution in 2.0 L, we need to use the formula:
Molarity (M) = moles of solute / volume of solution (L)
First, consider the given information:
Molarity (M) = 0.1M
Volume of solution = 2.0 L
Now, let's rearrange the formula to solve for moles of solute:
moles of solute = Molarity × volume of solution
moles of KI = 0.1M × 2.0 L = 0.2 moles
Next, we need to determine the molar mass of potassium iodide (KI). The molar mass of potassium (K) is approximately 39.10 grams/mole, and the molar mass of iodine (I) is approximately 126.90 grams/mole.
Molar mass of KI = (molar mass of potassium + molar mass of iodine)
= 39.10 g/mol + 126.90 g/mol
= 166.0 g/mol
Finally, we can calculate the grams of KI using the formula:
grams of KI = moles of KI × molar mass of KI
grams of KI = 0.2 moles × 166.0 g/mol
Therefore, to make a 0.1M solution in 2.0 L, you would need approximately 33.2 grams of potassium iodide.