what is the molarity of a solution prepared by dissolving 63.0 g NaF in enough water to make 250.0 mL of solution?
mols NaF = grams/molar mass
Then M = mols/L soln.
To calculate the molarity of a solution, you need to know the amount of solute (in moles) and the volume of the solution (in liters).
Given:
Mass of NaF = 63.0 g
Volume of solution = 250.0 mL (which is equivalent to 0.250 L)
Step 1: Calculate the number of moles of NaF.
To calculate moles, you need to divide the mass of NaF by its molar mass.
The molar mass of NaF (sodium fluoride) is:
Na (sodium) = 22.99 g/mol
F (fluorine) = 18.998 g/mol
So, the molar mass of NaF = 22.99 g/mol + 18.998 g/mol = 41.988 g/mol
Number of moles of NaF = Mass of NaF / Molar mass of NaF
= 63.0 g / 41.988 g/mol
≈ 1.5 mol (rounded to three significant figures)
Step 2: Convert the volume of the solution to liters.
Since molarity is defined as the number of moles of solute per liter of solution, we need to convert the volume from milliliters to liters.
Volume of solution in liters = 250.0 mL / 1000 mL/L
= 0.250 L
Step 3: Calculate the molarity of the solution.
Molarity = Number of moles of solute / Volume of solution in liters
Molarity = 1.5 mol / 0.250 L
= 6.0 mol/L
Therefore, the molarity of the solution prepared by dissolving 63.0 g NaF in enough water to make 250.0 mL of solution is 6.0 mol/L.