Question

You found the molarity of a solution containing 52 grams of sodium hydroxide. What would

the solution’s molarity be if you instead had 80.0 grams NaOH that was added to 200 milliliters
of water?
I need help with this (if you want more info on the molarity solution of sodium hydroxide the photo is the entire process I just need help with the NaOH that got added.)

Answer 1

the molarity of the solution would be 10.0 M if 80.0 grams of NaOH were added to 200 milliliters of water.

Step-by-step explanation:

To calculate the molarity of a solution, we can use the formula:

Molarity = moles of solute / volume of solution in liters

First, we need to calculate the moles of sodium hydroxide (NaOH) in the given mass:

moles NaOH = mass of NaOH / molar mass of NaOH

The molar mass of NaOH is 40.00 g/mol (the atomic mass of Na is 22.99 g/mol, the atomic mass of O is 15.99 g/mol, and the atomic mass of H is 1.01 g/mol).

For 52 g NaOH:

moles NaOH = 52 g / 40.00 g/mol = 1.30 mol

Now, we can calculate the molarity of the solution:

Molarity = moles NaOH / volume of solution in liters

We don't know the volume of the solution for the first case, so we cannot calculate the molarity.

For the second case, we have 80.0 g NaOH and 200 mL of water. We need to convert the volume to liters:

200 mL = 0.200 L

Now, we can calculate the moles of NaOH:

moles NaOH = 80.0 g / 40.00 g/mol = 2.00 mol

Finally, we can calculate the molarity of the solution:

Molarity = moles NaOH / volume of solution in liters

Molarity = 2.00 mol / 0.200 L

Molarity = 10.0 M

Therefore, the molarity of the solution would be 10.0 M if 80.0 grams of NaOH were added to 200 milliliters of water.