This question concerns quantifying solution concentration and molarity.

Question

Determine the mass of solute required to make each of the following solutions:

a. an 8.00 M solution of NaOH, with water added to bring the solution volume to 2.25 L

b. a 15.0 M solution of acetic acid, CH3COOH, with water added to bring the solution volume to 1.00 L.  

Answer 1

This question revolves around molarity, where M = moles per liter.

So, starting with part a), the first step for these problems is to find the number of moles of NaOH you need for the solution. We can find this by setting an equation of M = n/V, where M is the molarity, n is the moles of substance, and V is the volume in liters.

This gives us 8.00 mol/L = n / 2.25 L. Solving for n by multiplying 2.25 L on both sides gives us n = 18.0 moles of NaOH.

Now, we need to find what 18.0 moles of NaOH equates to in mass. To do so, we need to take the molar mass of NaOH (40.00 g/mol) and the moles and solve for the mass in grams. The molar mass was found by adding the molar masses of the individual components of the elements in the NaOH formula (22.99 g/mol for Na, 16.00 g/mol for O, and 1.01 g/mol for H).

Now, multiplying the molar mass by the amount of moles gives you 40.00 g/mol * 18 mol = 720. g NaOH, our final answer. Notice how I put the period at the end to keep 3 sig figs.

You can follow the same steps for part b) to get the answer of 901 g CH3COOH. I would encourage you to do the work on your own using these steps to make sure you can do it for future problems!

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