Question

To dilute a solution that is 40% saline, a chemist combines it with a solution that is 15% saline. How much of each solution should she use if she needs 150 mL of a solution that is 25% saline? Hint: solve by using system of equations

Answer 1

Answer:

60 ml of 40% saline and 90 ml of 15% saline

Step-by-step explanation:

We can call the amount of 40% solution x and the amount of 15% solution y.

x + y = 150 -- (1)

0.40x + 0.15y = 150 * 0.25 -- (2)  --- 150 * 0.25 = 37.5

40x + 15y = 3750 (Multiply (2) by 100 to get rid of decimals)

15x + 15y = 2250 -- (3) (Multiply (1) by 15)

25x = 1500           (Subtract (3) from (1)

x = 60

y = 150 - 60 = 90

Answer 2

Answer:

$37.5= 0.4 x +0.15 y$

We can solve for x and we got:

$x= \frac{37.5-0.15y}{0.4}= 93.75-0.375 y$

And replacing into the water condition we have:

$112.5 = (93.75-0.375 y)*0.6 +0.85y$

Solving for y we got:

$112.5= 56.25 -0.225 y+0.85 y$

$y = \frac{112.5-56.25}{0.625}= 90$

And then solving for x we got:

$x=\frac{37.5- 0.15*90}{0.4}= 60$

So we need 60 ml for the solution of 40% saline and 90 ml for the 15% saline solution

Step-by-step explanation:

We can solve this problem with the following system of equations:

$150*0.25 = x*0.4 + y *0.15$ salt

$150*(1-0.25)= x(1-0.4) +y(1-0.15)$ water

With x the ml of solution for 40% concentration and y the ml of solution at 15% of concentration

From the salt condition we have:

$37.5= 0.4 x +0.15 y$

We can solve for x and we got:

$x= \frac{37.5-0.15y}{0.4}= 93.75-0.375 y$

And replacing into the water condition we have:

$112.5 = (93.75-0.375 y)*0.6 +0.85y$

Solving for y we got:

$112.5= 56.25 -0.225 y+0.85 y$

$y = \frac{112.5-56.25}{0.625}= 90$

And then solving for x we got:

$x=\frac{37.5- 0.15*90}{0.4}= 60$

So we need 60 ml for the solution of 40% saline and 90 ml for the 15% saline solution