Serial Dilutions Practice

Dilution Problems
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  1. Serial Dilution Problems And Answers
  2. Serial Dilutions Practice Problems Medical Interventions
  3. Mi Serial Dilutions Practice Problems
  4. Serial Dilution Worksheet And Answers
  5. Serial Dilutions Practice Problems Pltw
  6. Serial Dilutions Practice Problems
  7. Serial Dilutions Practice Problems Pltw Answer Key

Problem #1: If you dilute 175 mL of a 1.6 M solution of LiCl to 1.0 L, determine the new concentration of the solution.

Solution:

  • A very quick lesson on how to do the math for serial and simple dilutions.
  • Dilutions Worksheet 1) If I add 25 mL of water to 125 mL of a 0.15 M NaOH solution, what will the molarity of the diluted solution be? 2) If I add water to 100 mL of a 0.15 M NaOH solution until the final volume is 150 mL, what will the molarity of the diluted solution be? 3) How much 0.05 M HCl solution can be made by diluting 250 mL of 10 M HCl?
M

But for what it's worth, the dilution factor for 1 mL stock + 99 mL water is: amount transferred / total amount = amount transferred / (amount transferred + amount water added) = 1 / (1+99) = 1/100 = 0.01. Below is an applet to practice finding dilution factors, and also to determine how much water to add to achieve a given dilution. Dilutions Worksheet 1) If I add 25 mL of water to 125 mL of a 0.15 M NaOH solution, what will the molarity of the diluted solution be? 2) If I add water to 100 mL of a 0.15 M NaOH solution until the final volume is 150 mL, what will the molarity of the diluted solution be? 3) How much 0.05 M HCl solution can be made by diluting 250 mL of 10 M HCl? WORKING DILUTION PROBLEMS It is a common practice to determine microbial counts for both liquid and solid specimens-suspensions of E. Coli in nutrient broth all the way to soil samples and hamburger meat. Most specimens have high enough numbers of microorganisms that the specimen has to be serially diluted to quantitate effectively.

1V1 = M2V2

(1.6 mol/L) (175 mL) = (x) (1000 mL)

x = 0.28 M

Note that 1000 mL was used rather than 1.0 L. Remember to keep the volume units consistent.

Problem #2: You need to make 10.0 L of 1.2 M KNO3. What molarity would the potassium nitrate solution need to be if you were to use only 2.5 L of it?

Solution:

M1V1 = M2V2

(x) (2.5 L) = (1.2 mol/L) (10.0 L)

x = 4.8 M

Please note how I use the molarity unit, mol/L, in the calculation rather than the molarity symbol, M.

Problem #3: How many milliliters of 5.0 M copper(II) sulfate solution must be added to 160 mL of water to achieve a 0.30 M copper(II) sulfate solution?

Solution:

M1V1 = M2V2

(5.00 mol/L) (x) = (0.3 mol/L) (160 + x)

5x = 48 + 0.3x

4.7x = 48

x = 10. mL (to two sig figs)

The solution to this problem assumes that the volumes are additive. That's the '160 + x' that is V2.

Problem #4: What volume of 4.50 M HCl can be made by mixing 5.65 M HCl with 250.0 mL of 3.55 M HCl?

Solution:

Here is the first way to solve this problem:

M1V1 + M2V2 = M3V3

Serial Dilution Problems And Answers

(3.55) (0.250) + (5.65) (x) = (4.50) (0.250 + x)

Where x is volume of 5.65 M HCl that is added
(0.250 + x) is total resultant volume

0.8875 + 5.65x = 1.125 + 4.50 x
1.15x = 0.2375
x= 0.2065 L

Total amount of 4.50 M HCl is then (0.250 + 0.2065) = 0.4565 L
Total amount = 456.5 mL

Here is the second way to solve this problem:

Since the amount of 5.65 M added is not asked for, there is no need to solve for it.

M1V1 + M2V2 = M3V3

(3.55) (250) + (5.65) (x - 250) = (4.50) (x)

That way, x is the answer you want, the final volume of the solution, rather than x being the amount of 5.65 M solution that is added.

Problem #5: A 40.0 mL volume of 1.80 M Fe(NO3)3 is mixed with 21.5 mL of 0.808M Fe(NO3)3 solution. Calculate the molar concentration of the final solution.

Solution:

Let's use a slightly different way to write the subscripts:

M1V1 + M2V2 = M3V3

There is no standard way to write the subscripts in problems of this type.

Substituting:

(1.80) (40.0) + (0.808) (21.5) = (M3) (40.0 + 21.5)

M3 = 1.45 M

Problem #6: To 2.00 L of 0.445 M HCl, you add 3.88 L of a second HCl solution of an unknown concentration. The resulting solution is 0.974 M. Assuming the volumes are additive, calculate the molarity of the second HCl solution.

Solution #1:

M1V1 + M2V2 = M

Serial Dilutions Practice Problems Medical Interventions

3V3

(0.445) (2.00) + (x) (3.88) = (0.974) (2.00 + 3.88)

0.890 + 3.88x = 5.72712

3.88x = 4.83712

x = 1.25 M (to three sig figs)

Solution #2:

1) Calculate moles HCl in 0.445 M solution:

(0.445 mol/L) (2.00 L) = 0.890 moles

2) Set up expression for moles of HCl in second solution:

(x) (3.88 L) = moles HCl in unknown solution

3) Calculate moles of HCl in final solution:

(0.974 mol/L) (5.88 L) = 5.73 moles

4) Moles of HCl in two mixed solutions = moles of HCl in final solution:

sure about because I wasted my haul hour to disable this pop-up. Set user settings to driver failed رفع مشکل.

0.890 moles + [(x) (3.88 L)] = 5.73 moles

x = 1.25 M (to three sig figs)

Problem #7: To what volume should you dilute 133 mL of an 7.90 M CuCl2 solution so that 51.5 mL of the diluted solution contains 4.49 g CuCl2?

Solution:

1) Find moles:

(4.49g CuCl2) (1 mole CuCl2 / 134.45 grams) = 0.033395 moles CuCl2

2) Find the molarity of the 51.5 mL of the diluted solution that contains 4.49g CuCl2:

(0.033395 moles CuCl2) / (0.0515 liters) = 0.648 M

3) Use the dilution formula:

M1V

Mi Serial Dilutions Practice Problems

1 = M2V2

(7.90 M) (133 mL) = (0.648 M) (V2)

V2 = 1620 mL

You should dilute the 133 mL of an 7.90 M CuCl2 solution to 1620 mL.

Problem #8: If volumes are additive and 95.0 mL of 0.55 M KBr is mixed with 165.0 mL of a BaBr2 solution to give a new solution in which [Br¯] is 0.65 M, what is the concentration of the BaBr2 used to make the new solution?

Solution:

moles of Br¯ from KBr: (0.55 mol/L) (0.095 L) = 0.05225 mol

moles of Br¯ in final solution: (0.65 mol/L) (0.260 L) = 0.169 mol

moles Br¯ provided by the BaBr2 solution: 0.169 - 0.05225 = 0.11675 mol

BaBr2 provides two Br¯ per formula unit so (0.11675 divided by 2) moles of BaBr2 are required for 0.11675 moles of Br¯ in the solution.

Serial Dilution Worksheet And Answers

molarity of BaBr2 solution: 0.058375 mol / 0.165 L = 0.35 M

Problem #9: 1.00 L of a solution is prepared by dissolving 125.6 g of NaF in it. If I took 180 mL of that solution and diluted it to 500 mL, determine the molarity of the resulting solution.

Solution:

1) Calculate moles of NaF:

125.6 g / 41.9 g/mol = 3.00 mol

2) Calculate moles in 180 mL of resulting solution:

Serial Dilutions Practice Problems Pltw

3.00 mol in 1000 mL so 3 x (180/1000) = 0.54 mol in 180 mL

3) Calculate molarity of diluted solution:

0.54 mol / 0.50 L = 1.08 mol/L = 1.08 M

Problem #10: What is the molar concentration of chloride ions in a solution prepared by mixing 100.0 mL of 2.0 M KCl with 50.0 mL of a 1.50 M CaCl2 solution?

(Warning: there's a complication in the solution. It has to do with the CaCl2.)

Serial Dilutions Practice Problems

Solution #1:

1) Get total moles of chloride:

KCl ⇒ (2.00 mol/L) (0.100 L) = 0.200 mol of chloride ion

CaCl2 ⇒ (1.50 mol/L) (0.0500 L) (2 ions / 1 formula unit) = 0.150 mol of chloride ion.

The '2 ions / 1 formula unit' is the problem child. The solution is 1.50 M in calcium chloride, but 3.00 M in just chloride ion.

total moles = 0.200 mol + 0.150 mol = 0.350 mol

2) Get chloride molarity:

0.350 mol / 0.150 L = 2.33 M

Solution #2:

Suppose you really wanted to use this equation:

M1V1 + M2V2 = M3V3

Set it up like this:

(2.00 mol/L) (0.100 L) + (3.00 mol/L) (0.0500 L) = (M3) (0.150 L)

Serial Dilutions Practice Problems Pltw Answer Key

Note that the CaCl2 molarity is 3.00 because that is the molarity of the solution from the point-of-view of the chloride ion.