Question

A 5.59 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 solution was determined to contain 5.70 ppm Ni.

Determine the molarity 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

a)

Get concentration of Ni in 100 mL

recall that ppm = mg of solute / liter of solution

5.70 ppm = mg of solute / (0.1 L)

mg = 5.7*0.1 = 0.57 mg = 0.57*10^-3 g of Nickel in 100 mL

mol of Ni = mass/MW = (0.57*10^-3) / 58.6934 = 0.00000971148 mol of Nickel

M = mol /V = (0.00000971148) /(0.1) = 0.0000971148 M

this is in the 100 mL aliquot so

M1*V1 = M2*V2

M1 = M2*V2/V1 = (0.0000971148)(100)/5

M1 = 0.001942296 mol of Nickel per liter, this is in the 5 mL aliquote

which is the same concentration than that of the 20 mL aliquot

b)

mass in original sample

M = 0.001942296 mol of Nickel per liter

we have only 20 mL

mol = MV = (0.001942296)(20*10^-3) = 0.00003884592 mol of Nickel in 20 mL / 5.59 g

mass = mol*MW = 0.00003884592*58.6934 = 0.002279 g of Nickel in sample

c)

% Ni = mass of Ni / Total mass * 100%

% Ni w/w = 0.002279 / 5.59*100% = 0.04076 %