Applications for Dilution Calculations Dilution is a process in which the concen

Question

Applications for Dilution Calculations Dilution is a process in which the concentration of a solution is lowered by adding more solvent Some of the reasons dilutions are performed are to minimze m when preparing a series of solutions at different concentrations, to save time and laboratory space, and to make more accurate measurements on an target concentration is very low. The new concentration of a diluted solution can be determined trom the following equation, sometimes called the dlution equation when the where Mi is the concentration of the initial solution, Vi is the volume of the initial solution, M2 is The terms Mi and M2 are often expressed in molanity, M, which is equal to i the notationsr Another way to wnite the equation is to replace subscripts of 1 and 2 with abbreviated Based on the inverse nature of this equivalency, small volumes of concentr which is in areas involving solution chemistry ated solutions can be used to prepare large volumes of diluted solutions with a known concentration, In order to understand how this equation is derived and why it holds true, the product of each side of the equation should be examined What value and unit do you get when you multiply a concentration of 0 580 M by a volume of 0 500 L ? This question can be expressed as 0,500 L 0,580 mol Express your answer to three significant figures with the appropriate units. Hints 12/2/2017

Explanation / Answer

PART A:

0.580 M x 0.500 L
= (0.580 mol/L) x 0.500 L                   ; Since M = mol/L)
= 0.290 mol

PART B:

Let the volume required for the 6.56 M stock solution be V1 mL

Since, we know that,

M1V1 = M2V2

6.56 M x V1 = 0.0316 M x 500 mL

V1 = (0.0316 M x 500 mL) / (6.56 M)
    = 2.408 mL
    = 2.41 mL

So, take 2.41 mL of 6.56 M stock solution and add solvent till it becomes 500 mL

PART C:

Let the concentration of the original sample be M1
Volume of the sample = 100 µL

The concentration of the diluted sample = 7.50 x 10-6 M
Volume of diluted sample = 2 mL + 100 µL
                                               = 2 x 1000 µL + 100 µL          ( 1 mL = 1000 µL)
                                               = 2000 µL + 100 µL         
                                               = 2100 µL

Since, we know that,

M1V1 = M2V2

M1 x 100 µL = (7.50 x 10-6) x 2100 µL

M1 = [ (7.50 x 10-6) x 2100 µL ] / (100 µL)
      = 0.0001575 µL
    = 0.000158 µL
     = 1.58 x 10-4 µL

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.