J.R. S. answered • 02/26/19

Ph.D. University Professor with 10+ years Tutoring Experience

V1M1 = V2M2

(25.00 ml)(5.00 M) = (100.00 ml)(x M)

x = 1.25 M

Without units the answer would be 1.25

Yess C.

asked • 02/26/19Quantitative dilution of solutions from higher to lower concentrations is a critical skill in this lab course. You have already diluted HCl from higher to lower concentrations using two different approaches requiring slightly different calculations

For example one procedure could use a 100.00 mL (V2) volumetric flask and a 10.00 mL (V1) volumetric pipet or pipettor to prepare a 0.500 M (M2) solution of HCl. A 10.00 mL volumetric pipet or pipettor is used to transfer 5.00 M HCl (M1) to a 100.00 mL (V2) volumetric flask and then the flask is filled exactly to 100.00 mL with distilled water and mixed well. The same dilution can be done using 5.00 mL (V1) and a 50.00 mL (V2) volumetric flask if less solution is needed.

**One way to make different concentrations of HCl (M2) in a 100.00 mL (V2) volumetric flask from a stock 5.00 M (M1) solution would be to use different volumetric pipets. What would be the molarity (M2) of the 100.00 mL (V2) of dilute HCl if you used a 25.00 (V1) mL volumetric pipet? Do not include units in your answer. **

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J.R. S. answered • 02/26/19

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Ph.D. University Professor with 10+ years Tutoring Experience

V1M1 = V2M2

(25.00 ml)(5.00 M) = (100.00 ml)(x M)

x = 1.25 M

Without units the answer would be 1.25

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