A 6.11 g sample of a solid containing Ni is dissolved in 20.0 mL water. A 5.00 mL aliquot of this solution is diluted to 100.0 mL and analyzed in the lab. The analyzed solution was determined to contain 5.14 ppm Ni . Determine the molar concentration of Ni in the 20.0 mL solution. concentration: 8.76 ×10 −3 M Determine the mass, in grams, of Ni in the original sample. mass: 0.0175 g Determine the weight percent of Ni in the sample. weight percent: 0.336 %

Respuesta :

Answer:

The molar concentrarion of Ni in the 20 mL solution is 1.75 x 10⁻³ mol/L.

The mass of Ni in the original sample is  0.0021 g.

The weight percent of Ni is 0.034%.

Explanation:

Molar concentration:

MW Ni = 58.6934 g / mol

5.14 ppm = 5.14 mg/L

5.14 mg Ni = 8.76 x 10⁻⁵ mol Ni

8.76 x 10⁻⁵ mol Ni ___ 1000 L

                x           ___  100 mL

                x = 8.76 x 10⁻⁶ mol Ni

8.76 x 10⁻⁶ mol Ni ____ 5 mL

            n               ____ 1000 mL

            n = 1.75 x 10⁻³  mol

The molar concentrarion of Ni in the 20 mL solution is 1.75 x 10⁻³ mol/L.

Mass in the original sample

1.75 x 10⁻³ mol Ni ___ 1000 mL

       y         ___  20 mL

       y = 3.50 x 10⁻⁵ mol Ni  = 0.0021 g Ni

The weigh percent:

( 0.0021 g / 6.11 g) x 100 = 0.034%

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