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If 10.0 mL of the solution on the right are withdrawn from the 100 mL beaker and diluted again in a similar manner, what is the new concentration?
0.0050 M NaCl
0.00050 M NaCl
0.050 M NaCl

If 100 mL of the solution on the right are withdrawn from the 100 mL beaker and diluted again in a similar manner what is the new concentration00050 M NaCl00005 class=

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duduburdfly

Answer: 0.050 M NaCl

Explanation:

So the original solution is diluted from a total volume of 10.0 mL to 100.0 mL.

The M represents molarity which is moles of NaCl divided by the total solution volume in liters.

So if our original volume were to be increased by a factor of 10 then the denominator would be changed.

Here is a mathematical illustration:

5/10 = .5

5/100 = .05

CintiaSalazar

Answer:

The new concentration is 0.050 M of NaCl

Explanation:

Dilution is a process in which a higher volume of solvent is added to a concentrated solution, this causes the concentration and volume of the resulting solution to be modified. So in a dilution the amount of solute does not vary.  What varies is the volume of the solvent: as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.

One way to calculate concentrations or volumes in dilutions is through the expression:

Ci * Vi = Cf * Vf

where

  • Ci = initial concentration
  • Vi = Initial volume
  • Cf = Final concentration
  • Vf = Final volume

In this case:

  • Ci= 0.5 M
  • Vi= 10 mL
  • Cf= ?
  • Vf= 100 mL

Replacing:

0.5 M*10 mL=Cf*100 mL

Solving:

[tex]Cf=\frac{0.5 M*10 mL}{100 mL}[/tex]

Cf= 0.050 M

The new concentration is 0.050 M of NaCl

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