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You ground your vitamin C tablet and dissolved the ground powder in 250. mL of water in a volumetric flask. The mass of the tablet was 1.0050 grams. The mass of the powder transferred into the volumetric flask was only 0.9875 grams. You used a 25.0 mL aliquot of this solution for your titration. If it takes 9.75 milliliters of your 0.0149M titrant, how many grams of ascorbic acid (MW 176.12) were in the original tablet? [enter the number as a decimal - do not use scientific notation. You must include leading zeros. Do not include units.]

Respuesta :

Answer:

0.2558 g

Explanation:

1.- In titration, since you have the concentration and the volume of the titrant and the aliquot that you take, you can use the formula C1 V1 = C2 V2.

2.- To obtain C2 = C1 V1  /V2

    Substitution C2 = (0.0149 M ) (9.75 ml) /25 ml = 0.0058 11 M

3.- This concentration is the number of moles that you would have in 1 L but you dissolve your tablet in 250 ml, so a rule of three.

4.- 0.005811 moles - 1 L

                     x        - 0.250 L

X= 0.250 L x 0.005811 moles / 1L = 0.00145275 moles

5.- using your molecular weight, you can obtain the grams with another rule of three.

0.00145275 moles -  x

                  1 mole - 176.12 g

x= 176.12 g  x 0.00145275 moles / 1 mole = 0.2558 g

Hope this was useful

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