contestada

Assume that you have 1, 2, 5, and 10 mL pipets and a 25 mL volumetric flask only. Describe in detail how you would prepare a solution with a concentration near 0.100 M from a 0.650 M stock solution using the glassware listed above. Calculate the final concentration of the diluted solution (it won't be exactly 0.100 M). It should be within 10% of 0.100 M. (Please see the Lab Skills for the accuracy on volumetric pipets and flasks.)

Respuesta :

Explanation:

The given data is as follows.

      [tex]V_{1}[/tex] = 25 ml,          [tex]M_{1}[/tex] = 0.10 M

      [tex]V_{2}[/tex] = x ml = [tex]25 ml \times \frac{0.001 L}{1 ml}[/tex],        [tex]M_{2}[/tex] = 0.650 M

It is known that relation between molarity and volume of two solutions is as follows.

        [tex]M_{1}V_{1}[/tex] = [tex]M_{1}V_{2}[/tex]

        [tex]0.10 M \times 0.025 L[/tex] = [tex]0.650 \times x[/tex]

                x = 0.00385 L

or,                = 3.85 ml

Since, this volume is closest to 4 ml so, we have to use a 2 ml pipette two times and then we have to put it in a 25 ml volumetric flask.

Hence, calculate the overall concentration as follows.

                    [tex]M_{1}V_{1}[/tex] = [tex]M_{1}V_{2}[/tex]

                     [tex]x \times 0.025 L[/tex] = [tex]0.650 M \times 0.004 L[/tex]    

                        x = 0.104 M

Therefore, it is about 3.8% error that is lower than the required error.

Otras preguntas