(a) A set of standards is run for a set of NiCl 2 •6H 2 O solutions. The data is

Question

(a)

A set of standards is run for a set of NiCl2•6H2O solutions. The data is plotted with absorbance on the y axis and the molarity of the nickel ion on the x-axis. The slope of the line is 5.16 M-1•cm-1 and the y-intercept is zero.

A sample of 1.50 g of NiSO4 with an unknown amount of water of hydration is dissolved in 25 mL of water and its absorbance is measured as 1.10. The pathlength was 1 cm

Calculate the concentration of nickel in the unknown sample.

(b)

Based on the answer to (a), calculate the mass of the nickel in the solid used in (a)

(c)

Based on the answer in (a) and (b), calculate the mass percent of the nickel in the original solid sample.

(d)

Based on the results in question (c), what is the most likely number of waters of hydration for NiSO4 from the following choices?

Explanation / Answer

Solution :-

part a )

A= e* b*c

1.10 = 5.16 M-1 cm-1 * 1 cm * c

1.10/5.16 M-1 cm-1 * 1 cm = c

0.2132 M = c

So the concentration of the Ni is 0.2132 M

(b)

Based on the answer to (a), calculate the mass of the nickel in the solid used in (a)

Solution :- moles = molarity * volume

                             =0.2132 mol per L * 0.025 L

                            = 0.00533 mol

Mass of Ni = moles * molar mass

                   = 0.00533 mol * 58.693 g per mol

                 = 03128 g Ni

So the mass of nickel in the sample is 0.3128 g

(c)

Based on the answer in (a) and (b), calculate the mass percent of the nickel in the original solid sample.

Solution :- mass % = (mass of Ni / mass of sample)*100%

                    = (0.3128 g / 1.50 g)*100%

                     = 20.85 %

So the mass % of the Ni is 20.85 %

Based on the results in question (c), what is the most likely number of waters of hydration for NiSO4 from the following choices?

Solution :- moles of Ni = 0.00533

So the moles of the NiSO4 = 0.00533 mol

Mass of NiSO4 = 0.00533 mol * 154.755 g per mol = 0.82484 g NiSO4

Therefore mass of water = 1.50 g – 0.82484 g = 0.67516 g

Moles of H2O = 0.67516 g / 18.0148 g per mol = 0.037478 mol

Now lets take the ratio of the moles of water to NiSO4

0.037478 mol water / 0.00533 mol NiSO4 = 7

So the moles of the water present in the hydrate are 7 as water of hydration.

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