For an experiment, 15 mL of 5% aq NaOH was added each time to the mixture. What is the molarity of 5% NaOH (solution density=1.055 g/mL)
To determine the molarity of a 5% NaOH solution, we need to first understand what is meant by a 5% solution.
A 5% solution means that there 5 grams of NaOH dissolved in a total of 100 mL of solution.
Since we have a density value for the solution, we can calculate the mass of NaOH present in 15 mL of the solution:
mass of solution = density * volume = 1.055 g/mL * 15 mL = 15.825 g
Now, we can calculate the mass of NaOH in the solution:
mass of NaOH = mass of solution * (5 g NaOH / 100 g solution) = 15.825 g * (5 g / 100 g) = 0.79125 g
The molarity (M) of a solution is given by the formula:
M = moles of solute / volume of solution in liters
To find the moles of NaOH in the solution, we need to know the molar mass of NaOH, which is 22.99 g/mol (Na) + 16.00 g/mol (O) + 1.01 g/mol (H) = 39.996 g/mol or approximately 40 g/mol.
moles of NaOH = mass of NaOH / molar mass of NaOH = 0.79125 g / 40 g/mol = 0.0198 mol
The volume of the solution is 15 mL or 0.015 L.
Molarity of the solution = 0.0198 mol / 0.015 L = 1.32 M
Therefore, the molarity of the 5% NaOH solution is approximately 1.32 M.