MHB Mixture Problem: Find Amount of Stronger Solution Needed

  • Thread starter Thread starter bergausstein
  • Start date Start date
  • Tags Tags
    Mixture
AI Thread Summary
A chemist is trying to determine how many gallons of a stronger salt solution (0.5 pounds per gallon) need to be added to 5 gallons of a weaker solution (0.2 pounds per gallon) to achieve a final concentration of 0.3 pounds per gallon. The equation derived from the problem is 1 + 0.5x = 0.3(x + 5), where x represents the gallons of the stronger solution added. The solution to this equation reveals that 2.5 gallons of the stronger solution is required. The discussion emphasizes the concept of weighted averages in mixture problems. The final concentration can be achieved by balancing the total salt content from both solutions.
bergausstein
Messages
191
Reaction score
0
A chemist has 5 gallons of salt solution with
a concentration of 0.2 pound per gallon and another solution
with a concentration of 0.5 pound per gallon.
How many gallons of the stronger solution
must be added to the weaker solution to get
a solution that contains 0.3 pound per gallon?

this my attempt

let $x=$ amount of stronger solution needed(in gallons)
$5-x=$ amount of weaker solution(in gallons)

$0.5(x)+0.2(5-x)=0.3(5)$

$x=$ 1.67 gallons

Is this correct? if not can you tell me why my method didn't work. thanks!
 
Last edited:
Mathematics news on Phys.org
We are going to add $x$ gallons of the stronger solution to the 5 gallons of weaker solution. So, we require, by equating two expressions for the total amount of salt:

$$5\cdot0.2+x\cdot0.5=(5+x)0.3$$

What do you get for $x$?

Do you see this is a weighted average?
 
MarkFL said:
We are going to add $x$ gallons of the stronger solution to the 5 gallons of weaker solution. So, we require, by equating two expressions for the total amount of salt:

$$5\cdot0.2+x\cdot0.5=(5+x)0.3$$

What do you get for $x$?

Do you see this is a weighted average?

that's stubborn weighted average again? :D:p

so there are two 5 gallons of mixture here? one with .2 lb/gal and the other .5lb/gal? Am I correct?

x = 2.5 gallons
 
Correct! :D

The problem you actually solved is how much of the weaker solution must be replaced by the stronger solution to get 5 gallons of a solution with 0.3 lb./gal concentration of salt. :D

bergausstein said:
...so there are two 5 gallons of mixture here? one with .2 lb/gal and the other .5lb/gal? Am I correct?...

You added this...we are not told how much of the stronger solution is available, but are left to assume that enough is available to get the desired solution.
 
:D good
 
Hello, bergausstein!

Here is my approach to mixture problems.

A chemist has two solutions.
Solution A: concentration of 0.2 pound of salt per gallon .
Solution B: concentration of 0.5 pound per gallon.
How many gallons of solution B must be added to 5 gallons of solution A to get a solution that contains 0.3 pound per gallon?
He has 5 gallons of A which has 0.2 pound salt per gallon.
This contains: .(0.2)(5) = 1 pound of salt.

He adds x gallons of B which has 0.5 pound salt per gallon.
This contains: .0.5x pounds of salt.

Hence, the mixture will contain 1 + 0.5x pounds of salt.But we know that the mixture has x +5 gallons
. . which has 0.3 pounds of salt per gallon.
This contains 0.3(x+5) pounds of salt.We just described the final amount of salt in two ways.

There is our equation! . . . 1 + 0.5x \;=\;0.3(x+5)
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top