chemistry question

Question

A  5.80g 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  5.46ppm Ni.

Determine the molar concentration of Ni in the 20.0 mL solution.

Determine the mass, in grams, of Ni, in the original sample.

Determine the weight percent of Ni in the sample.

 

Answer 1

5.46 ppm Ni = 5.46 mg/liter

Now to correct for aliquoting:

5.00 ml from 20.0 ml = 4x

5 mls diluted to 100.0 mls = 20 x

Factor = 4 x 20 = 80 x

5.46 mg/liter x 80 = 436.8 mg/L

Concentration of Ni in the original 20.0 ml sample = 436.8 mg/L x 1g/1000 mg x 1 mol Ni/58.69 g = 0.00744 moles/L = 0.00744 M

Mass of Ni (in grams) in original 20.0 ml (0.020 L) sample =436.8 mg/L x 0.020 L x 1 g/1000 mg = 0.00874 g

Weight percent of Ni in the sample = mass of Ni/total mass (x100%) = 0.00874 g/5.80 g (x100%) = 0.151%