QUESTION 9 The next three questions focus on a 26 ppm Ag+(aq) solution that Kemm

Question

QUESTION 9 The next three questions focus on a 26 ppm Ag+(aq) solution that Kemmi and Doc are using. First, calculate the molarity of the silver cation (Ag+) in the solution. Assume the solution density is the same as pure water. (2 significant digits, unit M) 1 points

QUESTION 10 Kemmi dilutes the 26 ppm Ag+(aq) solution by twice pipetting 15 mL of solution into a 100 mL volumetric flask. She fills the remaining volume with water. Use M1V1 = M2V2 with M1 equal to 26 ppm. Solve for the diluted concentration (M2) in units of ppm. (2 significant digits, unit ppm) 1 points

QUESTION 11 Doc wants to make a 10.4 ppm Ag+(aq) solution from the original 26 ppm solution. He has a 25 mL volumetric flask available. Use M1V1 = M2V2 with the M1 and M2 values in units of ppm, to predict which volumetric pipet Doc should use to make his dilution. (2 significant digits, unit mL)

Explanation / Answer

9) The concentration of Ag+ in the solution is 26 ppm = 26 mg/L.

Molar mass of Ag = 107.8 g/mol.

Therefore, moles of Ag = (26 mg)*(1 g/1000 mg)*(1 mole/107.8 g) = 0.00024119 mole.

Molar concentration of Ag+ = (0.00024119 mole/1 L) = 0.00024119 mole/L = 2.4119*10-4 mole/L 2.4*10-4 M (ans).

10) Given M1 = 26 ppm, V1 = 15 mL and V2 = 100 mL; plug in the equation to obtain M2 as

(26 ppm)*(15 mL) = M2*(100 mL)

===> M2 = (26*15)/(100) ppm = 3.9 ppm.

The concentration of Ag+ in the dilute solution is 3.9 ppm (ans).

11) We are given M1 = 26 ppm, M2 = 10.4 ppm and V2 = 25 mL; we are required to find out V1. Plug in the values and obtain

(26 ppm)*V1 = (10.4 ppm)*(25 mL)

===> V1 = (10.4*25)/(26) mL = 10 mL.

The volume of the original solution required = 10 mL (ans).

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