Chemistry Review

Question

What vlume of a 0350 M potassium carbonate solution could be prepared starting with 50.0 g of the pure solid?

Answer 1

Hi Nicole! The work for this problem would look like this:

(50.0g) x (1 mol / 138.21g) x (1 L / 0.350 mol) = 1.03 L

So, the answer to this equation would be 1.03 L. Now, a few quick notes before you copy down this work:

1. I assumed you meant to say 0.350 M and not 0350 M but if I am mistaken, then just change it in the work I provided for you and plug it into your calculator to get the final answer.
2. The (1 mol / 138.21g) is the molar mass for Potassium Carbonate (K₂CO₃) just incase you're wondering where I got that from.
3. I'm not sure what units your answer needs to be in, so I left it in the standard unit of liters (L) but feel free to change it/convert it to something else depending on what the question is asking for.

Hope this helps! Please feel free to reach out if you have any other questions or help needed.