How would you prepare 3.5 L of a 0.9M solution of KCl?
A. Add 23 g of KCl to a 3.5 L container; then add enough water to dissolve the KCl and fill the container to the 3.5 L mark.
B. Add 233 g of KCl to a 3.5 L container; then add enough water to dissolve the KCl and fill the container to the 3.5 L mark.
C. Add 287 g of KCl to a 3.5 L container; then add enough water to dissolve the KCl and fill the container to the 3.5 L mark.
D. Add 567 g of KCl to a 3.5 L container; then add enough water to dissolve the KCl and fill the container to the 3.5 L mark.
mass=Molarity*molemas*volume
= .9*74.5*3.5= you do it.
Then add water to the 3.5Liter mark, stir, and label.
To determine the correct answer, we need to calculate the amount of KCl needed to prepare a 0.9M solution in a 3.5 L container.
The molarity of a solution is defined as the number of moles of solute per liter of solution. Therefore, to determine the number of moles of KCl needed, we can use the formula:
moles of solute = molarity × volume of solution (in liters)
We are given that the desired volume of the solution is 3.5 L and the desired molarity is 0.9M.
Substituting these values into the formula:
moles of KCl = 0.9M × 3.5 L = 3.15 moles
Now, we need to calculate the molar mass of KCl. Potassium (K) has a molar mass of approximately 39.1 g/mol, and chlorine (Cl) has a molar mass of approximately 35.5 g/mol. Therefore, the molar mass of KCl is:
molar mass of KCl = (39.1 g/mol) + (35.5 g/mol) = 74.6 g/mol
Finally, to determine the mass of KCl needed, we multiply the number of moles of KCl by its molar mass:
mass of KCl = moles of KCl × molar mass of KCl = 3.15 moles × 74.6 g/mol = 234.99 g
Rounding to the nearest whole number, we find that approximately 235 g of KCl is needed to prepare a 0.9M solution in a 3.5 L container.
Among the given options:
A. Add 23 g of KCl to a 3.5 L container; then add enough water to dissolve the KCl and fill the container to the 3.5 L mark.
B. Add 233 g of KCl to a 3.5 L container; then add enough water to dissolve the KCl and fill the container to the 3.5 L mark.
C. Add 287 g of KCl to a 3.5 L container; then add enough water to dissolve the KCl and fill the container to the 3.5 L mark.
D. Add 567 g of KCl to a 3.5 L container; then add enough water to dissolve the KCl and fill the container to the 3.5 L mark.
Option B is the correct answer since it requires adding 233 g of KCl to a 3.5 L container, which is the closest value to the calculated 235 g of KCl needed.