Question #d35f2

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Question #d35f2

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Answer 1 · 2015-02-18T22:13:04

The student should use 1.0 × 10³ µL of clarified yeast extract.

The first question is, How much 100 % ethanol is needed to make 1.3 mL of 26 % ethanol?

This is a standard dilution calculation.

$Volume = 1.3 mL \times \frac{26 %}{100 %} = 0.34 mL$ $"Volume" = "1.3 mL" × "26 %"/"100 %" = "0.34 mL"$

So you need to add enough clarified extract to make a total of 1.3 mL of solution.

How much extract is needed?

If there is no change of volume on mixing,

$V_{extract} = V_{total} - V_{ethanol} = 1.3 mL - 0.34 mL = 1.0 mL = 1.0 \times 10^{3} µL$ $V_"extract" = V_"total" – V_"ethanol" = "1.3 mL" – "0.34 mL" = "1.0 mL" = 1.0 × 10^3"µL"$

Note: The answer can have only 2 significant figures, because that is all you gave for the volume of the sample. If you need more precision, you will have to recalculate.

Answer 2 · 2015-02-18T22:16:03

You would need $962$ $962$ $μ L$ $muL$ of clarified yeast extract.

Here's how you'd go about solving this problem. You know that you must mix ethanol and clarified yeast extract together to get a $1.3-mL$ $"1.3-mL"$ sample that is $26%$ $"26%"$ ethanol.

Now, I'm assuming you're dealing with a $26% v/v$ $"26% v/v"$ solution, which is defined as

$% volume = \frac{V_{solute}}{V_{solution}} \cdot 100$ $"% volume" = V_("solute")/V_("solution") * 100$

If ethanol is your solute, a $26% v/v$ $"26% v/v"$ solution would imply that every $100 mL$ $"100 mL"$ of solution contain $26 mL$ $"26 mL"$ of ethanol. As a result, your sample will contain

$1.3 mL solution \cdot \frac{26 mL ethanol}{100 mL solution} = 0.338 mL ethanol$ $"1.3 mL solution" * ("26 mL ethanol")/("100 mL solution") = "0.338 mL ethanol"$

This means that the volume of clarified yeast extract must be

$V_{solution} = V_{ethanol} + V_{yeast} \Rightarrow V_{yeast} = V_{solution} - V_{ethanol}$ $V_("solution") = V_("ethanol") + V_("yeast") => V_("yeast") = V_("solution") - V_("ethanol")$

$V_{yeast} = 1.3 mL - 0.338 mL = 0.962 mL$ $V_("yeast") = "1.3 mL" - "0.338 mL" = "0.962 mL"$ , or

$0.962 mL \cdot \frac{1000 μ L}{1 mL} = 962$ $"0.962 mL" * (1000 muL)/("1 mL") = "962"$ $μ L$ $muL$

Rounded to two sig figs, the answer should be

$V_{yeast} = 960$ $V_("yeast") = "960$ $μ L$ $muL$

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Question #d35f2

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