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If 4.5ml of yeast prep + 0.5ml of 0.25 M glucose is separated into 2ml in 2 similar flasks. How many micromoles of glucose are there in one flask?

Using M1V1=M2V2 (0.5ml/1000)*0.25M = M2 * 2ml = 0.0000625M

Using n=cv n= 0.0000625*2 = 0.000125moles =125umoles

Is this correct?

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  • $\begingroup$ No, V2 should be 5.0 mL $\endgroup$ – A.K. Mar 15 '19 at 13:58
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Use of $M_1V_1 = M_2V_2$ is a good start, but you lost at the same place. Identifying three known values out of four is important. You have initial volume ($V_1$) and concentration ($M_1$) of your glucose solution, which are $\pu{0.50 mL}$ and $\pu{0.25 M} = \pu{0.25 mol\:L^{-1}}$, respectively. You also knows the final volume ($V_2$) after dilution, which is $\pu{(4.5+0.5) mL}=\pu{5.0 mL}$ (See A.K.'s comment). Now you see, your unknown is $M_2$. From rearranging the equation $M_1V_1 = M_2V_2$ for $M_2$, you get:

$$M_2 = \frac {M_1V_1}{V_2} = \frac{\pu{0.25 mol\:L^{-1}} \times \pu{0.50 mL}}{\pu{5.0 mL}} = \pu{0.025 mol\:L^{-1}}$$

Now you can use similar approach to find amount of glucose in $\pu{2.0 mL}$ of $\pu{0.025 M}$ glucose+yeast solution:

$$\text{amount of glucose in}\; \pu{2.0 mL}= \pu{0.025 mol\:L^{-1}} \times \pu{2.0 mL}\times \frac{\pu{1.0 L}}{\pu{10^3 mL}}\times \frac{\pu{10^6 \mu mol}}{\pu{1.0 mol}}=\pu{50 \mu mol}$$

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No.

0.25M = 0.25 mmol/ml 0.5 ml of this contains 0.125 mmol, and diluting it to 5 ml (4.5+0.5ml) gives a concentration of 0.025 mmol/ml. Transfering 2 ml of this dilution to a flask then gives 0.05 mmol (=50 umol) glucose in the flask.

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