Mixture Problem: Find Amount of Stronger Solution Needed

  • Context: MHB 
  • Thread starter Thread starter bergausstein
  • Start date Start date
  • Tags Tags
    Mixture
Click For Summary
SUMMARY

The discussion centers on solving a mixture problem involving two salt solutions: one with a concentration of 0.2 pounds per gallon and another with 0.5 pounds per gallon. The goal is to determine how many gallons of the stronger solution (0.5 pounds per gallon) must be added to 5 gallons of the weaker solution (0.2 pounds per gallon) to achieve a final concentration of 0.3 pounds per gallon. The correct calculation yields that 2.5 gallons of the stronger solution is required to reach the desired concentration.

PREREQUISITES
  • Understanding of basic algebraic equations
  • Knowledge of concentration calculations
  • Familiarity with weighted averages
  • Concept of mixture problems in chemistry
NEXT STEPS
  • Study algebraic manipulation techniques for solving equations
  • Learn about concentration and dilution in chemistry
  • Explore additional mixture problem examples
  • Review weighted average calculations in various contexts
USEFUL FOR

Chemistry students, educators, and anyone interested in solving mixture problems or enhancing their understanding of concentration calculations.

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)
 

Similar threads

Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
3
Views
4K
  • · Replies 19 ·
Replies
19
Views
8K