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.]

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

i25.9

chronic ischemic heart disease, unspecified

i25.9 is a billable/specific icd-10-cm code that can be used to indicate a diagnosis for reimbursement purposes.

explanation: