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MIDTERM_Gravimetry_Spectroscopy

Total questions: 65

Worksheet time: 5hrs 54mins

Name
Class
Date
1.

Which gravimetric method involves the separation of the analyte from a solution of the sample as a precipitate or conversion to a compound of known composition that can be weighed?

a)

Precipitation gravimetry

b)

Volatilization gravimetry

c)

Particulate gravimetry

d)

Electrogravimetry

e)

Gravimetric titrimetry

2.

Which gravimetric method involves the separation of the analyte by deposition on an electrode by an electrical current? The mass of the product then provides a measure of the analyte concentration.

a)

Precipitation gravimetry

b)

Volatilization gravimetry

c)

Particulate gravimetry

d)

Electrogravimetry

e)

Gravimetric titrimetry

3.

Which gravimetric method involves the loss of a volatile species that gives rise to an analytical signal which can be used to calculate the amount of analyte in a sample?

a)

Precipitation gravimetry

b)

Volatilization gravimetry

c)

Particulate gravimetry

d)

Electrogravimetry

e)

Gravimetric titrimetry

4.

Which gravimetric method involves determining the mass of a particulate analyte after its separation from the matrix? (E.g. Total suspended solids in environmental monitoring)

a)

Precipitation gravimetry

b)

Volatilization gravimetry

c)

Particulate gravimetry

d)

Electrogravimetry

e)

Gravimetric titrimetry

5.

Who discovered gravimetric analysis?

a)

Robert Bunsen

b)

Theodore Richards

c)

Erwin Schrödinger

d)

Niels Bohr

e)

Gustav Kirchoff

6.

This type of co-precipitation involves impurities of similar ions to that of the precipitate taking part of the internal structure of the precipitate's crystal lattice.

a)

Occlusion

b)

Inclusion

c)

Surface adsorption

d)

Mechanical entrapment

e)

Peptization

7.

This type of co-precipitation involves impurities trapped inside the pockets of the internal structure of the precipitate's crystal lattice.

a)

Occlusion

b)

Inclusion

c)

Surface adsorption

d)

Mechanical entrapment

e)

Peptization

8.

This type of co-precipitation involves impurities adhering to the surface of the precipitate.

a)

Occlusion

b)

Inclusion

c)

Surface adsorption

d)

Mechanical entrapment

e)

Peptization

9.

When two or more precipitate crytals grow together and eventually merging, a part of the solution is trapped inside the growing crystals resulting in a type of co-precipitation known as...

a)

Occlusion

b)

Inclusion

c)

Surface adsorption

d)

Mechanical entrapment

e)

Peptization

10.

Choose ALL the conditions that are best for achieving an ideal precipitate:

a)

Faster rate of nucleation

b)

Slower rate of nucleation

c)

Faster rate of particle growth

d)

Slower rate of particle growth

e)

Equal rate of nucleation and particle growth

11.

Choose ALL the conditions that are best for achieving an ideal precipitate:

a)

Faster rate of nucleation

b)

Slower rate of nucleation

c)

Faster rate of particle growth

d)

Slower rate of particle growth

e)

Equal rate of nucleation and particle growth

12.

Choose ALL the conditions that are best for achieving an ideal precipitate:

a)

Low supersaturation

b)

High supersaturation

c)

Low temperature

d)

High temperature

e)

Dilute solution

13.

Choose ALL the conditions that are best for achieving an ideal precipitate:

a)

Low supersaturation

b)

High supersaturation

c)

Low temperature

d)

High temperature

e)

Dilute solution

14.

Choose ALL the conditions that are best for achieving an ideal precipitate:

a)

Ksp value of the precipitate is high.

b)

Ksp value of the precipitate is low.

c)

Solubility of the precipitate is high.

d)

Solubility of the precipitate is low.

e)

The resulting precipitate has known structure/identity.

15.

Choose ALL the conditions that are best for achieving an ideal precipitate:

a)

Big particle size.

b)

Small particle size.

c)

Formation of electric bilayer.

d)

Digestion

e)

Drying of the precipitate.

16.

Choose ALL the conditions that are best for achieving an ideal precipitate:

a)

Slow addition of the precipitating agent.

b)

Quick addition of the precipitating agent.

c)

Vigorous stirring after addition of precipitating agent.

d)

Gentle stirring after addition of precipitating agent.

e)

Using high temperature when adding the precipitating agent.

17.

The following are the properties of a colloid:

a)

Scatters light (Tyndall effect).

b)

Constantly suspended particles moving in a solution (Brownian motion).

c)

Passes through the filter paper.

d)

Filtered by the filter paper.

e)

Small particle size.

18.

The following results to the formation of (suspended) colloidal precipitates:

a)

Formation of an electric bilayer

b)

Peptization

c)

Digestion

d)

Presence of supporting electrolytes

e)

Washing the precipitate with boiling water

19.

A mixture of unknown salts containing a sulfate salt has been weighed to be 3.00 g. After dissolving the solution in 500 mL of water, heating it, and adding hot barium chloride (40.0 mL, 0.500 M); the precipitate has been collected after succeeding digestion, and the mass was found to be 0.325 g. What is the percentage (by mass) of sulfur in the mixture?

a)

4.05%

b)

2.98%

c)

5.00%

d)

1.49%

e)

3.24%

20.

A mixture of unknown salts containing a sulfate salt has been weighed to be 3.00 g. After dissolving the solution in 500 mL of water, heating it, and adding hot barium chloride (40.0 mL, 0.500 M); the precipitate has been collected after succeeding digestion, and the mass was found to be 0.325 g. What is the percentage (by mass) of sulfur in the mixture?

a)

4.05%

b)

2.98%

c)

5.00%

d)

1.49%

e)

3.24%

21.

A mixture of unknown salts containing a sulfate salt has been weighed to be 3.00 g. After dissolving the solution in 500 mL of water, heating it, and adding hot barium chloride (40.0 mL, 0.500 M); the precipitate has been collected after succeeding digestion, and the mass was found to be 0.325 g. What is the percentage (by mass) of sulfur in the mixture?

a)

4.05%

b)

2.98%

c)

5.00%

d)

1.49%

e)

3.24%

22.

A 0.2356 g sample containing only NaCl (58.45 g/mol) and BaCl2 (208.23 g/mol) yielded 0.4637 of dried AgCl (143.32 g/mol) when reacted with excess silver nitrate. Calculate the percent of each salt (NaCl and BaCl2, respectively) in the sample.

a)

40.50%, 59.50%

b)

29.80%, 70.20%

c)

5.000%, 95.00%

d)

33.40%, 66.60%

e)

55.01%, 44.99%

23.

A 0.2356 g sample containing only NaCl (58.45 g/mol) and BaCl2 (208.23 g/mol) yielded 0.4637 of dried AgCl (143.32 g/mol) when reacted with excess silver nitrate. Calculate the percent of each salt (NaCl and BaCl2, respectively) in the sample.

a)

40.50%, 59.50%

b)

29.80%, 70.20%

c)

5.000%, 95.00%

d)

33.40%, 66.60%

e)

55.01%, 44.99%

24.

A 0.2356 g sample containing only NaCl (58.45 g/mol) and BaCl2 (208.23 g/mol) yielded 0.4637 of dried AgCl (143.32 g/mol) when reacted with excess silver nitrate. Calculate the percent of each salt (NaCl and BaCl2, respectively) in the sample.

a)

40.50%, 59.50%

b)

29.80%, 70.20%

c)

5.000%, 95.00%

d)

33.40%, 66.60%

e)

55.01%, 44.99%

25.

A 0.2356 g sample containing only NaCl (58.45 g/mol) and BaCl2 (208.23 g/mol) yielded 0.4637 of dried AgCl (143.32 g/mol) when reacted with excess silver nitrate. Calculate the percent of each salt (NaCl and BaCl2, respectively) in the sample.

a)

40.50%, 59.50%

b)

29.80%, 70.20%

c)

5.000%, 95.00%

d)

33.40%, 66.60%

e)

55.01%, 44.99%

26.

An iron ore was analyzed by dissolving 0.5662-g sample in concentrated HCl. The resulting solution was diluted with water, and the iron(III) was precipitated as the hydrous oxide Fe2O3 · xH2O by the addition of NH3. After filtration and washing, the residue was ignited at a high temperature to give 0.5394 g of pure Fe2O3 (159.69 g/mol). Calculate the % Fe (55.847 g/mol) in the sample.

a)

33.32%

b)

66.63%

c)

35.45%

d)

13.46%

e)

3.378%

27.

An iron ore was analyzed by dissolving 0.5662-g sample in concentrated HCl. The resulting solution was diluted with water, and the iron(III) was precipitated as the hydrous oxide Fe2O3 · xH2O by the addition of NH3. After filtration and washing, the residue was ignited at a high temperature to give 0.5394 g of pure Fe2O3 (159.69 g/mol). Calculate the % Fe (55.847 g/mol) in the sample.

a)

33.32%

b)

66.63%

c)

35.45%

d)

13.46%

e)

3.378%

28.

When calculating for the mass of the analyte given the mass of the precipitate by gravimetric factor method, what is the gravimetric factor if: precipitate=PbO2, analyte=Pb3O4?

a)

3 MM Pb3O4/ MM PbO2

b)

MM Pb3O4/ 3 MM PbO2

c)

MM Pb3O4/ MM PbO2

d)

3 MM PbO2/ MM Pb3O4

e)

MM PbO2/ MM Pb3O4

29.

When calculating for the mass of the analyte given the mass of the precipitate by gravimetric factor method, what is the gravimetric factor if: precipitate=Mg2P2O7, analyte=Mg?

a)

2 MM Mg2P2O7/ MM Mg

b)

MM Mg/ 2 MM Mg2P2O7

c)

MM Mg2P2O7/ MM Mg

d)

2 MM Mg/ MM Mg2P2O7

e)

MM Mg2P2O7/ 2 MM Mg

30.

Which of the following refers to the sum of the suspended solids and dissolved solids?

a)

TSS

b)

TDS

c)

TS

d)

FSS

e)

FDS

31.

Which of the following refers to the filterable solids in a water sample?

a)

TSS

b)

TDS

c)

TS

d)

FSS

e)

FDS

32.

Which of the following refers to the non-filterable solids dissolved in a water sample?

a)

TSS

b)

TDS

c)

TS

d)

FSS

e)

FDS

33.

Refer to the given data. Calculate the TSS expressed in ppm.

(a)  

34.

Refer to the given data. Calculate the TSS expressed in ppm.

(a)  

35.

Refer to the given data. Calculate the VSS expressed in ppm.

(a)  

36.

Refer to the given data. Calculate the VSS expressed in ppm.

(a)  

37.

Which of the following refers to a form of energy having a wave-particle duality that is transmitted through space at enormous velocities?

a)

Heat

b)

Sound waves

c)

Electromagnetic radiation

d)

Electrical energy

e)

Magnetic field

38.

This term refers to how much light passes through a sample unchanged (i.e. not absorbed, deflected, reflected, or refracted).

a)

Transmittance

b)

Absorbance

c)

Molar absorption

d)

Molar absorptivity extinction coefficient

e)

Beer's Law

39.

This term refers to the measure of how strongly a chemical species/analyte absorbs light at a particular wavelength. This is an intrinsic property and differs per analyte.

a)

Transmittance

b)

Absorbance

c)

Molar absorption

d)

Molar absorptivity extinction coefficient

e)

Beer's Law

40.

What is the wavelength at which the analyte exhibits maximum absorbance?

a)

λ

b)

λmax

c)

λmin

d)

510 nm

e)

A

41.

Refer to the given figure. What is the maximum absorption wavelength of caffeine?

a)

450 nm

b)

275 nm

c)

300 nm

d)

200 nm

e)

510 nm

42.

Refer to the given diagram of a spectrophotometer. Which part of the spectrophotometer narrows down light to a single wavelength?

a)

Light source

b)

Monochromator

c)

Cuvette

d)

Photoresistor

e)

Amplifier

43.

Refer to the given diagram of a spectrophotometer. Which part of the spectrophotometer detects the light transmitted?

a)

Light source

b)

Monochromator

c)

Cuvette

d)

Photoresistor

e)

Amplifier

44.

Refer to the given diagram of a spectrophotometer. Which part of the spectrophotometer enhances the electrical current detected by the photoresistor?

a)

Light source

b)

Monochromator

c)

Cuvette

d)

Photoresistor

e)

Amplifier

45.

Refer to the given diagram of a spectrophotometer. Which part of the spectrophotometer holds the sample solution?

a)

Light source

b)

Monochromator

c)

Cuvette

d)

Photoresistor

e)

Amplifier

46.

Choose ALL the limitations of UV-Vis spectrophotometry from the given choices:

a)

Cannot be used for colored solutions

b)

Cannot be used for very concentrated analyte solutions (5.00+ M levels)

c)

Cannot be used for very diluted analyte solutions (ppb levels)

d)

Cannot be used for transparent solutions

e)

Requires dilution or serial dilution for concentrated analyte solutions

47.

Which term describes the direct proportionality of absorbance to the concentration of analyte?

a)

Absorbance

b)

Transmittance

c)

Beer-Lambert Law

d)

Electromagnetic spectrum

e)

λmax

48.

Convert the 1.250 absorbance to %Transmittance.

a)

56.23%

b)

17.75%

c)

34.87%

d)

5.623%

e)

1.775%

49.

Convert the 1.250 absorbance to %Transmittance.

a)

56.23%

b)

17.75%

c)

34.87%

d)

5.623%

e)

1.775%

50.

Convert 45.0% transmittance to absorbance.

a)

0.584

b)

0.347

c)

0.461

d)

0.857

e)

0.184

51.

Convert 45.0% transmittance to absorbance.

a)

0.584

b)

0.347

c)

0.461

d)

0.857

e)

0.184

52.

Calculate the molar absorptivity extinction coefficient (ε) for a 45.0 ppm analyte with 0.956 absorbance, in a 1.00 cm wide cuvette. (A = εbc)

a)

0.154 ppm^(-1) cm^(-1)

b)

0.0352 ppm^(-1) cm^(-1)

c)

0.0212 ppm^(-1) cm^(-1)

d)

0.0653 ppm^(-1) cm^(-1)

e)

0.412 ppm^(-1) cm^(-1)

53.

Calculate the molar absorptivity extinction coefficient (ε) for a 45.0 ppm analyte with 0.956 absorbance, in a 1.00 cm wide cuvette. (A = εbc)

a)

0.154 ppm^(-1) cm^(-1)

b)

0.0352 ppm^(-1) cm^(-1)

c)

0.0212 ppm^(-1) cm^(-1)

d)

0.0653 ppm^(-1) cm^(-1)

e)

0.412 ppm^(-1) cm^(-1)

54.

One common way to determine phosphorus in urine is to treat the sample after removing the protein with molybdenum (VI) and then reducing the resulting 12-molybdophosphate complex with ascorbic acid to give an intense blue-colored species called molybdenum blue. The absorbance of molybdenum blue can be measured at 650 nm. A 24-hour urine sample was collected, and the patient produced 1122 mL in 24 hours. A 1.00 mL aliquot of the sample was treated with Mo(VI) and ascorbic acid and diluted to a volume of 50.00 mL. A calibration curve was prepared by treating 1.00 mL aliquots of phosphate standard solutions in the same manner as the urine sample. The absorbances of the standards and the urine sample were obtained at 650 nm and the following results obtained below. What is the equation of the corresponding Beer's law calibration curve?

a)

y = 0.2454x + 0.0654

b)

y = 0.2056x + 0.024

c)

y = 0.4523x + 0.0125

d)

y = 0.6521x + 0.09843

e)

y = 0.5291x – 0.03512

55.

One common way to determine phosphorus in urine is to treat the sample after removing the protein with molybdenum (VI) and then reducing the resulting 12-molybdophosphate complex with ascorbic acid to give an intense blue-colored species called molybdenum blue. The absorbance of molybdenum blue can be measured at 650 nm. A 24-hour urine sample was collected, and the patient produced 1122 mL in 24 hours. A 1.00 mL aliquot of the sample was treated with Mo(VI) and ascorbic acid and diluted to a volume of 50.00 mL. A calibration curve was prepared by treating 1.00 mL aliquots of phosphate standard solutions in the same manner as the urine sample. The absorbances of the standards and the urine sample were obtained at 650 nm and the following results obtained below. What is the equation of the corresponding Beer's law calibration curve?

a)

y = 0.2454x + 0.0654

b)

y = 0.2056x + 0.024

c)

y = 0.4523x + 0.0125

d)

y = 0.6521x + 0.09843

e)

y = 0.5291x – 0.03512

56.

One common way to determine phosphorus in urine is to treat the sample after removing the protein with molybdenum (VI) and then reducing the resulting 12-molybdophosphate complex with ascorbic acid to give an intense blue-colored species called molybdenum blue. The absorbance of molybdenum blue can be measured at 650 nm. A 24-hour urine sample was collected, and the patient produced 1122 mL in 24 hours. A 1.00 mL aliquot of the sample was treated with Mo(VI) and ascorbic acid and diluted to a volume of 50.00 mL. A calibration curve was prepared by treating 1.00 mL aliquots of phosphate standard solutions in the same manner as the urine sample. The absorbances of the standards and the urine sample were obtained at 650 nm and the following results obtained below. What is the equation of the corresponding Beer's law calibration curve?

a)

y = 0.2454x + 0.0654

b)

y = 0.2056x + 0.024

c)

y = 0.4523x + 0.0125

d)

y = 0.6521x + 0.09843

e)

y = 0.5291x – 0.03512

57.

One common way to determine phosphorus in urine is to treat the sample after removing the protein with molybdenum (VI) and then reducing the resulting 12-molybdophosphate complex with ascorbic acid to give an intense blue-colored species called molybdenum blue. The absorbance of molybdenum blue can be measured at 650 nm. A 24-hour urine sample was collected, and the patient produced 1122 mL in 24 hours. A 1.00 mL aliquot of the sample was treated with Mo(VI) and ascorbic acid and diluted to a volume of 50.00 mL. A calibration curve was prepared by treating 1.00 mL aliquots of phosphate standard solutions in the same manner as the urine sample. The absorbances of the standards and the urine sample were obtained at 650 nm and the following results obtained below. What is the phosphate concentration in the diluted sample subjected to spectroscopy?

a)

2.45 ppm

b)

3.25 ppm

c)

1.15 ppm

d)

1.79 ppm

e)

2.40 ppm

58.

One common way to determine phosphorus in urine is to treat the sample after removing the protein with molybdenum (VI) and then reducing the resulting 12-molybdophosphate complex with ascorbic acid to give an intense blue-colored species called molybdenum blue. The absorbance of molybdenum blue can be measured at 650 nm. A 24-hour urine sample was collected, and the patient produced 1122 mL in 24 hours. A 1.00 mL aliquot of the sample was treated with Mo(VI) and ascorbic acid and diluted to a volume of 50.00 mL. A calibration curve was prepared by treating 1.00 mL aliquots of phosphate standard solutions in the same manner as the urine sample. The absorbances of the standards and the urine sample were obtained at 650 nm and the following results obtained below. What is the phosphate concentration in the diluted sample subjected to spectroscopy?

a)

2.45 ppm

b)

3.25 ppm

c)

1.15 ppm

d)

1.79 ppm

e)

2.40 ppm

59.

One common way to determine phosphorus in urine is to treat the sample after removing the protein with molybdenum (VI) and then reducing the resulting 12-molybdophosphate complex with ascorbic acid to give an intense blue-colored species called molybdenum blue. The absorbance of molybdenum blue can be measured at 650 nm. A 24-hour urine sample was collected, and the patient produced 1122 mL in 24 hours. A 1.00 mL aliquot of the sample was treated with Mo(VI) and ascorbic acid and diluted to a volume of 50.00 mL. A calibration curve was prepared by treating 1.00 mL aliquots of phosphate standard solutions in the same manner as the urine sample. The absorbances of the standards and the urine sample were obtained at 650 nm and the following results obtained below. What is the phosphate concentration in the diluted sample subjected to spectroscopy?

a)

2.45 ppm

b)

3.25 ppm

c)

1.15 ppm

d)

1.79 ppm

e)

2.40 ppm

60.

One common way to determine phosphorus in urine is to treat the sample after removing the protein with molybdenum (VI) and then reducing the resulting 12-molybdophosphate complex with ascorbic acid to give an intense blue-colored species called molybdenum blue. The absorbance of molybdenum blue can be measured at 650 nm. A 24-hour urine sample was collected, and the patient produced 1122 mL in 24 hours. A 1.00 mL aliquot of the sample was treated with Mo(VI) and ascorbic acid and diluted to a volume of 50.00 mL. A calibration curve was prepared by treating 1.00 mL aliquots of phosphate standard solutions in the same manner as the urine sample. The absorbances of the standards and the urine sample were obtained at 650 nm and the following results obtained below. What is the phosphate concentration in the original undiluted sample?

a)

245 ppm

b)

325 ppm

c)

115 ppm

d)

240 ppm

e)

120 ppm

61.

One common way to determine phosphorus in urine is to treat the sample after removing the protein with molybdenum (VI) and then reducing the resulting 12-molybdophosphate complex with ascorbic acid to give an intense blue-colored species called molybdenum blue. The absorbance of molybdenum blue can be measured at 650 nm. A 24-hour urine sample was collected, and the patient produced 1122 mL in 24 hours. A 1.00 mL aliquot of the sample was treated with Mo(VI) and ascorbic acid and diluted to a volume of 50.00 mL. A calibration curve was prepared by treating 1.00 mL aliquots of phosphate standard solutions in the same manner as the urine sample. The absorbances of the standards and the urine sample were obtained at 650 nm and the following results obtained below. What is the phosphate concentration in the original undiluted sample?

a)

245 ppm

b)

325 ppm

c)

115 ppm

d)

240 ppm

e)

120 ppm

62.

One common way to determine phosphorus in urine is to treat the sample after removing the protein with molybdenum (VI) and then reducing the resulting 12-molybdophosphate complex with ascorbic acid to give an intense blue-colored species called molybdenum blue. The absorbance of molybdenum blue can be measured at 650 nm. A 24-hour urine sample was collected, and the patient produced 1122 mL in 24 hours. A 1.00 mL aliquot of the sample was treated with Mo(VI) and ascorbic acid and diluted to a volume of 50.00 mL. A calibration curve was prepared by treating 1.00 mL aliquots of phosphate standard solutions in the same manner as the urine sample. The absorbances of the standards and the urine sample were obtained at 650 nm and the following results obtained below. What is the phosphate concentration in the original undiluted sample?

a)

245 ppm

b)

325 ppm

c)

115 ppm

d)

240 ppm

e)

120 ppm

63.

One common way to determine phosphorus in urine is to treat the sample after removing the protein with molybdenum (VI) and then reducing the resulting 12-molybdophosphate complex with ascorbic acid to give an intense blue-colored species called molybdenum blue. The absorbance of molybdenum blue can be measured at 650 nm. A 24-hour urine sample was collected, and the patient produced 1122 mL in 24 hours. A 1.00 mL aliquot of the sample was treated with Mo(VI) and ascorbic acid and diluted to a volume of 50.00 mL. A calibration curve was prepared by treating 1.00 mL aliquots of phosphate standard solutions in the same manner as the urine sample. The absorbances of the standards and the urine sample were obtained at 650 nm and the following results obtained below. If the absorbance of the treated urine sample is 1.25, speculate about the concentration of the sample, the reliability of results, and what could have been done.

a)

The concentration of the sample is below that of the standard solutions (0.596 ppm), therefore, the result is not reliable. The a different dilution should have been done.

b)

The concentration of the sample is above that of the standard solutions (5.96 ppm), therefore, the result is not reliable. The treated sample should have been further diluted with a 1/2 dilution factor.

c)

The concentration of the sample is below that of the standard solutions (5.96 ppm), therefore, the result is not reliable. The treated sample should have been further diluted with a 1/2 dilution factor.

d)

The concentration of the sample is above that of the standard solutions (0.596 ppm), therefore, the result is not reliable. The a different dilution should have been done.

e)

The concentration of the sample is below that of the standard solutions (4.96 ppm), therefore, the result is not reliable. The treated sample should have been further diluted with a 1/2 dilution factor.

64.

One common way to determine phosphorus in urine is to treat the sample after removing the protein with molybdenum (VI) and then reducing the resulting 12-molybdophosphate complex with ascorbic acid to give an intense blue-colored species called molybdenum blue. The absorbance of molybdenum blue can be measured at 650 nm. A 24-hour urine sample was collected, and the patient produced 1122 mL in 24 hours. A 1.00 mL aliquot of the sample was treated with Mo(VI) and ascorbic acid and diluted to a volume of 50.00 mL. A calibration curve was prepared by treating 1.00 mL aliquots of phosphate standard solutions in the same manner as the urine sample. The absorbances of the standards and the urine sample were obtained at 650 nm and the following results obtained below. If the absorbance of the treated urine sample is 1.25, speculate about the concentration of the sample, the reliability of results, and what could have been done.

a)

The concentration of the sample is below that of the standard solutions (0.596 ppm), therefore, the result is not reliable. The a different dilution should have been done.

b)

The concentration of the sample is above that of the standard solutions (5.96 ppm), therefore, the result is not reliable. The treated sample should have been further diluted with a 1/2 dilution factor.

c)

The concentration of the sample is below that of the standard solutions (5.96 ppm), therefore, the result is not reliable. The treated sample should have been further diluted with a 1/2 dilution factor.

d)

The concentration of the sample is above that of the standard solutions (0.596 ppm), therefore, the result is not reliable. The a different dilution should have been done.

e)

The concentration of the sample is below that of the standard solutions (4.96 ppm), therefore, the result is not reliable. The treated sample should have been further diluted with a 1/2 dilution factor.

65.

One common way to determine phosphorus in urine is to treat the sample after removing the protein with molybdenum (VI) and then reducing the resulting 12-molybdophosphate complex with ascorbic acid to give an intense blue-colored species called molybdenum blue. The absorbance of molybdenum blue can be measured at 650 nm. A 24-hour urine sample was collected, and the patient produced 1122 mL in 24 hours. A 1.00 mL aliquot of the sample was treated with Mo(VI) and ascorbic acid and diluted to a volume of 50.00 mL. A calibration curve was prepared by treating 1.00 mL aliquots of phosphate standard solutions in the same manner as the urine sample. The absorbances of the standards and the urine sample were obtained at 650 nm and the following results obtained below. If the absorbance of the treated urine sample is 1.25, speculate about the concentration of the sample, the reliability of results, and what could have been done.

a)

The concentration of the sample is below that of the standard solutions (0.596 ppm), therefore, the result is not reliable. The a different dilution should have been done.

b)

The concentration of the sample is above that of the standard solutions (5.96 ppm), therefore, the result is not reliable. The treated sample should have been further diluted with a 1/2 dilution factor.

c)

The concentration of the sample is below that of the standard solutions (5.96 ppm), therefore, the result is not reliable. The treated sample should have been further diluted with a 1/2 dilution factor.

d)

The concentration of the sample is above that of the standard solutions (0.596 ppm), therefore, the result is not reliable. The a different dilution should have been done.

e)

The concentration of the sample is below that of the standard solutions (4.96 ppm), therefore, the result is not reliable. The treated sample should have been further diluted with a 1/2 dilution factor.