Commercial Sodium hydroxide has a concentration of 19.4mol/L. How many milliliters of sodium hydroxide solution can be prepared from 59g or sodium hydroxide.
Your post just isn't clear. I can prepare almost any number of mL from 59 g NaOH but of WHAT concentration. Do you mean to start with 59 g NaOH and ask how many mL of 19.4M NaOH can be prepared from that? If so then
mols NaOH in 59 g = grams/molar mass = 59/40 = 1.475 mols NaOH.
Then M NaOH = mols NaOH/L NaOH
19.4 = 1.475/L and solve for L the convert to mL.
To solve this problem, we need to use the molar mass of sodium hydroxide (NaOH) and the given concentration to calculate the number of moles of NaOH in 59 grams. Then, we can use the molarity to find the volume of solution in milliliters.
Step 1: Calculate the molar mass of NaOH.
The atomic mass of Na (sodium) is 22.99 g/mol, the atomic mass of O (oxygen) is 16.00 g/mol, and the atomic mass of H (hydrogen) is 1.01 g/mol.
So, the molar mass of NaOH = (22.99 g/mol) + (16.00 g/mol) + (1.01 g/mol) = 40.00 g/mol.
Step 2: Convert the given mass of NaOH to moles.
Number of moles = mass / molar mass
Number of moles = 59 g / 40.00 g/mol = 1.48 moles.
Step 3: Use the concentration to find the volume of the solution in milliliters.
Molarity (M) = moles of solute / volume of solution (in liters)
19.4 mol/L = 1.48 moles / volume (in liters)
Now, we can rearrange the formula to solve for the volume:
volume (in liters) = moles of solute / Molarity
volume (in liters) = 1.48 moles / 19.4 mol/L ≈ 0.076 L.
Step 4: Convert the volume from liters to milliliters.
1 L = 1000 mL, so we multiply the volume in liters by 1000 to get the volume in milliliters.
0.076 L x 1000 mL/L = 76 mL.
Therefore, 59 grams of sodium hydroxide can be used to prepare approximately 76 milliliters of sodium hydroxide solution.