### Help needed for molarity problem

Posted:

**Fri Oct 14, 2016 8:15 pm**Hello, I am not sure how to approach this problem. Can someone please help me?

Created by Dr. Laurence Lavelle

https://lavelle.chem.ucla.edu/forum/

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=10&t=15945

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Posted: **Fri Oct 14, 2016 8:15 pm**

Hello, I am not sure how to approach this problem. Can someone please help me?

Posted: **Fri Oct 14, 2016 8:52 pm**

In order to solve this problem, you will be using MV=MV. You are given the initial mass of the solute, the initial volume, the final volume, and you are asked to find the final molarity. The molar mass of KCl is 74.551 g/mol (found using the molar masses of K and Cl). First, find the initial molarity of the solution:

M initial = 55.1g KCl x ((1mol KCl)/(74.551 g KCl)) x (1/.075 L) = 9.85455147 mol/L

This allows you to cancel out grams, and you are left with mol/L.

Next, rearrange the formula MV=MV to solve for M final. You will get M = MV/V. Plug in and solve:

M final = ((9.85455147 mol/L)(.075 L))/(.125 L) = 5.912730882 mol/L = 5.91 mol/L --- use 3 sig figs because that was given initially.

M initial = 55.1g KCl x ((1mol KCl)/(74.551 g KCl)) x (1/.075 L) = 9.85455147 mol/L

This allows you to cancel out grams, and you are left with mol/L.

Next, rearrange the formula MV=MV to solve for M final. You will get M = MV/V. Plug in and solve:

M final = ((9.85455147 mol/L)(.075 L))/(.125 L) = 5.912730882 mol/L = 5.91 mol/L --- use 3 sig figs because that was given initially.