math concentration

Question

A KCl saline solution has a concentration of 11.1 g/mL. If you have 80 mL of it, and need to dilute it down to 6 g/mL.

You are required to dilute it with a weaker concentration of 2.7 g/mL, how much of that weaker solution must you add?

Answer 1

Hello, thank you for taking the time to post your question!

The key on this type of question is to set up an algebraic expression that accurately captures all of the variables and given information. So on this one we have

(11.1 x 80)  +  (2.7 x V2)  =  6 x (80 + V2) for the equation, where V2 represents the volume of the weaker solution that needs to be added

Solving that for V2 I get

888 + 2.7*V2 = 480 + 6*V2

408 = 3.3*V2

V2 = 123.64

So we need to add about 123.64 mL of the weaker solution to get the total mixture

I hope that helps get you moving in the right direction! Feel free to reach out if you have any additional question beyond that :)

Answer 2

here's one way to look at it , solve it

y axis is grams, x-axis is ml

have point (80,888)

need to get to point (x,y) where slope =y/x=6 (g/ml)

From (80,888) establish line to (x,y) with slope 2.7 (g/ml)

point and slope ,,,(y-888)/(x-80)=2.7

substitute y=6x

now have

6x-888=2.7(x-80)

x=203.6 ml is the total amount of liquid

203.6-80=123.6 ml is the amount of 2.7g/ml solution to add

check,,,80*11.1+123.6*2.7=1221.7 total grams

80+123.6=203.6 total ml

1221.7/203.6=6.0 g/ml