Question

A 6.04 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 6.88 ppm Ni . Determine the molar concentration of Ni in the 20.0 mL solution.

*concentration= ? M

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

*mass= ? g

Determine the weight percent of Ni in the sample.

*weight percent= ? %

Answer 1

Answer:-

Given:-

mass of the Ni sample = 6.04 g

volume of the Ni sample = 20.0 mL

concentration of Ni in 100 mL solution (C₁) = 6.88 ppm

volume of diluted solution (V₁) = 100.0 mL

volume of aliquot of the Ni sample (V₂) = 5.00 mL

concentration of Ni in aliquot of the Ni sample (C₂) = ?

molar concentration of Ni in the 20.0 mL solution = ?

mass of Ni in the sample = ?

weight percent of Ni in the sample = ?

According to the formula

C₁V₁ = C₂V₂

C₂ = C₁V₁ / V₂

So

concentration of Ni in aliquot of the Ni sample (C₂) = concentration of Ni in 100 mL solution (C₁) $\times$ volume of diluted solution (V₁) / volume of aliquot of the Ni sample (V₂)

concentration of Ni in aliquot of the Ni sample (C₂) = 6.88 ppm $\times$ 100.0 mL / 5.00 mL

concentration of Ni in aliquot of the Ni sample (C₂) = 137.6 ppm

As we know that

5.00 mL volume of aliquot of the Ni sample contained = 137.6 ppm concentration of Ni

1.00 mL volume of aliquot of the Ni sample contained = 137.6 ppm / 5.00  concentration of Ni

then

20.0 mL volume of the Ni sample contained = 137.6 ppm $\times$ 20.0 / 5.00  concentration of Ni

20.0 mL volume of the Ni sample contained = 550.4‬0 ppm concentration of Ni

therefore

concentration of Ni in the 20.0 mL sample = 550.4‬0 ppm concentration of Ni

since we know that

ppm of the component = mass of component (g) $\times$ 10⁶ / volume of solution (mL)

So

ppm of the Ni = mass of Ni (g) $\times$ 10⁶ / volume of solution (mL)

mass of Ni (g) = ppm of the Ni $\times$ volume of solution (mL) / 10⁶

mass of Ni (g) = 550.4‬0 ppm  $\times$ 20.0 mL / 10⁶

mass of Ni (g) = 11008 / 10⁶

mass of Ni in the original sample (g) = 11008 $\times$ 10^-6 g (i.e the answer)

mass of Ni in the original sample (g) = 0.011008‬ g (i.e the answer)

As we know that

molar mass of Ni = 58.69 g / mol

volume of sample solution = 20.0 mL = 0.02 L

Also we know that

molarity of compound = wt. of compound / molar mass of compound $\times$ volume of solution (L)

molarity of Ni or molar concentration of Ni in the 20.0 mL solution = wt. of Ni / molar mass of Ni $\times$ volume of solution (L)

molarity of Ni or molar concentration of Ni in the 20.0 mL solution = 0.011008‬ g / 58.69 g / mol $\times$ 0.02 L

molarity of Ni or molar concentration of Ni in the 20.0 mL solution = 0.011008‬ mol / 1.1738‬ L

molarity of Ni or molar concentration of Ni in the 20.0 mL solution = 9.4 $\times$ 10^-3‬ mol / L ( i.e the answer)

molarity of Ni or molar concentration of Ni in the 20.0 mL solution = 9.4 $\times$ 10^-3 M ( i.e the answer)

As we know that according to the formula

weight percent (%) of Ni = wt. of Ni (g)  $\times$ 100 / wt. of sample (g)

weight percent (%) of Ni = 0.011008‬ g $\times$ 100 / 6.04 g

weight percent (%) of Ni = 1.1008‬ / 6.04

weight percent (%) of Ni = 0.182 % ( i.e the answer)