Units on Fundamental G#19

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Briana Perez 3A
Posts: 35
Joined: Fri Sep 28, 2018 12:23 am

Units on Fundamental G#19

Postby Briana Perez 3A » Thu Oct 04, 2018 9:26 pm

Edition 7. Fundamental G. #19 "A sample of 1.345 M K2SO4 of volume 12.56 mL is diluted to 250.0 mL. What is the molar concentration of K2SO4 in the diluted solution?

I understand how to solve the problem by multiplying (1.345 mol L)(0.01256 L) / 0.2500 L. The answer comes out to be 0.06757 mol L-1

However, I do not understand why the L is to the negative 1 power.

Can someone please explain it to me.

Kathryn 1F
Posts: 66
Joined: Fri Sep 28, 2018 12:19 am

Re: Units on Fundamental G#19

Postby Kathryn 1F » Thu Oct 04, 2018 9:44 pm

Hi!
The units of molarity are moles per litre, which can also be written as mol/L or mol*L^-1, because raising a number to a negative power places it on the denominator. This can be figured out using the equation

M=(Moles of solute)/(Liters of solution), with the units being moles/l

You can also figure out the units by cancelling them out in your chain equation.


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