# Help required for the calculation of moles [closed]

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?

• No, V2 should be 5.0 mL – A.K. Mar 15 '19 at 13:58

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