Suppose you were instructed to add 0.2 mL of sample to 9.8 mL of diluent, but instead added 2.0 mL of sample. What was the intended dilution, and what was the actual dilution?

Question

Biology

Francis Vega

21 December, 22:55

Suppose you were instructed to add 0.2 mL of sample to 9.8 mL of diluent, but instead added 2.0 mL of sample. What was the intended dilution, and what was the actual dilution?

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Answers (2)

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Santos Barr · Answer 1

Intended dilution = 50 x; Actual dilution = 5.9 x

Explanation:

Dilution factor = (Sample Volume + Diluent Volume) / Sample Volume

Intended dilution factor = (0.2 + 9.8) / 0.2

= 50

Actual dilution factor = (2 + 9.8) / 2

= 5.9

Ray Dudley · Answer 2

Intended dilution: 50 times

Actual dilution: 5.9 times

Explanation:

Dilution: To decrease the concentration of the solute in a particular solution

Dilution = total volume of solution/volume of sample

Here, actual sample: 0.2ml, diluent: 2 ml

So, intended dilution = 10/0.2 = 50 times

Actual sample: 2 ml

So, actual dilution=11.8/2 = 5.9 times