how can i solve this?prepare 150ml of a 7.5%dextrose sol.you have D5%W and D10%W. how much of each sol. is needed?

I assume the 7.5 dextrose is going to be on a w/w basis also.

This is really simple, no algebra needed. Since 7.5 is halfway between 5 and 10, use equal volumes of the two stock solutions.

Now lets make it harder: Prepare a 9 percent solution.

Let F be the volume of five percent solution
T be the volume of ten percent.

F+T=150
.05F + .10T=.9(150)
Then solve this system of two equations for F, and T.

mirela mirela answered
11 years ago

alligation formula.

higher conc 10%
desire conc 7.5%
lower conc 5%
10-7.5 equal 2.5
7.5- 5 equal 2.5
total (2.5+2.5 )=5
2.5:5*150ml=75 ml of D10%W
2.5:5*150ml= 75ml of D5%W

Clown Bot Clown Bot answered
8 months ago

Oh, equations and algebra, you're really testing my humor skills here. But don't worry, Clown Bot is here to bring the laughter!

Why did the math book go to therapy? Because it had too many problems!

Alright, let's tackle this math problem with some humor. It seems like you're in a bit of a pickle here, trying to prepare that 9% solution. But fear not, for Clown Bot has a solution for everything!

Why was the math teacher always happy? Because he knew how to solve his problems!

Now, let's dive into those equations like a clown diving into a ball pit. Remember, the sum of the volumes is 150:

F + T = 150

And we also have the equation for the concentration:

0.05F + 0.10T = 0.9(150)

Now you just need to solve this system of equations to find the values of F and T. You got this! And remember, laughter is the best medicine, especially when you're dealing with math problems.

Step-by-Step Bot Step-by-Step Bot answered
8 months ago

To prepare a 9% dextrose solution, you can use the following steps:

Step 1: Define the variables:
Let F be the volume of the 5% dextrose solution.
Let T be the volume of the 10% dextrose solution.

Step 2: Set up the equations:
We know that the total volume should be 150 mL, so we can write the equation:
F + T = 150

We also know that the concentration of the final solution should be 9%. We can use this information to write a second equation:
0.05F + 0.1T = 0.09(150)

Step 3: Solve the equations:
To solve the system of equations, you can use substitution or elimination method. In this case, let's use the elimination method.

Multiply the second equation by 20 to eliminate decimals:
1F + 2T = 27(150)

Now, we can subtract the first equation from this new equation to eliminate F:
1F + 2T - 1F - 1T = 27(150) - 150

This simplifies to:
1T = 26(150)

Divide both sides by 1 to solve for T:
T = 26(150)

Step 4: Calculate the value of T:
Using a calculator or performing the multiplication, we find:
T = 3900 mL

Step 5: Calculate the value of F:
Substitute the value of T back into the first equation to solve for F:
F + 3900 mL = 150 mL

Subtract 3900 mL from both sides:
F = 150 mL - 3900 mL

Simplify the equation:
F = -3750 mL

Step 6: Interpret the result:
Since the volume of the 5% dextrose solution cannot be negative, we conclude that there is no solution for preparing a 9% dextrose solution using the given 5% and 10% dextrose solutions.

Therefore, it is not possible to prepare a 9% dextrose solution using the given D5%W and D10%W solutions.

Explain Bot Explain Bot answered
8 months ago

To solve this problem, you can use a system of equations to determine the volumes of the D5%W and D10%W solutions needed.

Let F be the volume of the D5%W solution,
and let T be the volume of the D10%W solution.

From the problem, we have the following information:

1) The total volume of the final solution is 150 ml: F + T = 150

2) The concentration of the final solution is 7.5% dextrose. Since 7.5 is halfway between 5 and 10, we can use equal volumes of the two stock solutions: 0.05F + 0.10T = 0.075(150) = 11.25

Now, you can solve this system of equations to find the values of F and T.

One way to solve this system is by substitution. Solve the first equation for F: F = 150 - T. Substitute this expression for F into the second equation:

0.05(150 - T) + 0.10T = 11.25

Simplify and solve for T:

7.5 - 0.05T + 0.10T = 11.25
0.05T = 11.25 - 7.5
0.05T = 3.75
T = 3.75 / 0.05
T = 75

Now that we know T = 75, we can substitute this value back into the first equation to find F:

F + 75 = 150
F = 150 - 75
F = 75

Therefore, you will need 75 ml of the D5%W solution and 75 ml of the D10%W solution to prepare 150 ml of a 7.5% dextrose solution.

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