If there are 0.625 g of NaCl left in a beaker that originally contained 75.0 mL of saltwater, what must have been the molar concentration of the original solution? The molar mass of NaCl is 58.44 g/mol.
Didn't I do this for you last night? At your original post?
mols NaCl = grams/molar mass
Then M = mols/L
To find the molar concentration of the original solution, we need to calculate the number of moles of NaCl in the beaker and then divide it by the volume of the solution.
First, let's calculate the number of moles of NaCl left in the beaker.
Number of moles = Mass of NaCl / Molar mass of NaCl
Given that the mass of NaCl remaining in the beaker is 0.625 g and the molar mass of NaCl is 58.44 g/mol:
Number of moles = 0.625 g / 58.44 g/mol
Number of moles = 0.0107 mol (rounded to four decimal places)
Next, we need to determine the volume of the original solution. The volume of the original solution is given as 75.0 mL.
Finally, we can calculate the molar concentration (also known as the molarity) of the original solution.
Molarity (M) = Number of moles / Volume of solution (in liters)
Since the volume of the solution is given in milliliters, we need to convert it to liters.
Volume of solution = 75.0 mL = 75.0 mL * (1 L / 1000 mL) = 0.075 L
Now we can calculate the molar concentration:
Molarity (M) = 0.0107 mol / 0.075 L
Molarity (M) = 0.143 M (rounded to three decimal places)
Therefore, the molar concentration of the original solution must have been 0.143 M.