## Post- Assessment Problem 16

Anokhi Patel 2B
Posts: 76
Joined: Fri Aug 09, 2019 12:17 am

### Post- Assessment Problem 16

How would I solve this problem, I'm not really sure where to start or how to implement the equation.

16. A solution is prepared by dissolving 55.1 g of KCl in approximately 75 mL of water and then adding water to a final volume of 125 mL. What is the molarity of KCl(aq) in this solution?

A. 0.276 mol.L-1

B. 5.91 mol.L-1

C. 98.5 mol.L-1

D. 3.70 mol.L-1

Thank you!

Natalie Wang 1B
Posts: 51
Joined: Sat Sep 14, 2019 12:15 am
Been upvoted: 1 time

### Re: Post- Assessment Problem 16

To find molarity, you have to divide the mols of solute by the volume of the solution. Convert KCL to mols. Then, convert the 125 mL to L. Finally, divide the mols of KCL by the volume in L to get the answer.

Posts: 53
Joined: Thu Jul 25, 2019 12:17 am

### Re: Post- Assessment Problem 16

Well, first you'd use the calculated molar mass of KCl (74.5513 g/mol) to convert the mass of KCl into moles of KCl:
5.51 g KCl * 1 mol KCl/74.5513 g KCl = 0.739 mol KCl
Then, the problem states that water is added to a final volume of 125 mL. Basically this means that water was poured into the container (already containing 75 mL of water) until a total volume of 125 mL was found within it. (I also found this statement confusing!). So, all you have to do now to find the molarity is to divide the moles of KCl (0.739) by the total volume of water in the container (125 mL = 0.125 L).
Thus, the molarity of KCl in this solution = 0.739 mol/125 mL = 0.739 mol/0.125 L = 5.91 mol * L-1. The answer is option B. Hope this helps! Please reach out if you need further clarification regarding my explanation.