Question

In: Chemistry

Question 1) A 1.0 mL aliquot (sample) is taken from a 1.0L solution. Which of the...

Question

1) A 1.0 mL aliquot (sample) is taken from a 1.0L solution. Which of the following is/are true when comparing the 1.0 mL solution to the 1.0L solution? Choose all that are true.

-The molarity of the 1.0 mL solution is less than the molarity of the 1.0L solution

-The molarity of each is the same

-The number of moles of solute in each solution is the same

-The number of moles of solute in the 1.0mL solution is less than the number of moles of solute in the 1.0 L solution

2) 5.00 mL is withdrawn from a 1.00L solution of NaCl and diluted to 100.00 mL. If the concentration of the NaCl in the 100.00 mL solution is 0.033, what is the concentration of NaCl in the original 1.00 L solution?

3) What is used as a blank (reference/zero absorbance) for the analysis of the samples in the spectrophotometer?

-Deioniozed water

- Nothing-air

-Acetylsalicylic acid standard solution

- Iron(III) chloride solution

4) The acetylsalycylic acid - iron (III) complex appears violet in color. What color of light is used to measure the absorbance of the solution?

- red

- orange

- green

- blue

- violet

5) A calibration curve for absorbance versus the concentration of the chemical indicator crystal violet is found to have a slope of 48. If a sample of unknown concentration has an absorbance of 1.38, what is the concentration in the unknown sample?

Solutions

Expert Solution

1) A 1.0 mL aliquot (sample) is taken from a 1.0L solution. True statements when comparing the 1.0 mL solution to the 1.0L solution are,

-The molarity of each is the same

-The number of moles of solute in the 1.0mL solution is less than the number of moles of solute in the 1.0 L solution

---

2) 5.00 mL is withdrawn from a 1.00L solution of NaCl and diluted to 100.00 mL. If the concentration of the NaCl in the 100.00 mL solution is 0.033,

the concentration of NaCl in the original 1.00 L solution =0.033 M x 100 ml/5 ml = 0.66 M

--

3) Blank (reference/zero absorbance) for the analysis of the samples in the spectrophotometer is,

- Deioniozed water

- Iron(III) chloride solution

---

4) The acetylsalycylic acid - iron (III) complex appears violet in color. The color of light is used to measure the absorbance of the solution is,

- orange

- blue

--

5) A calibration curve for absorbance versus the concentration of the chemical indicator crystal violet is found to have a slope of 48. If a sample of unknown concentration has an absorbance of 1.38,

the concentration in the unknown sample = 1.38/48 = 0.03 M

queen_honey_blossom answered 10 months ago
Next > < Previous

Related Solutions

A 10.010.0 mL sample of whiskey was diluted to 500.0 mL. A 4.004.00 mL aliquot of...
A 10.010.0 mL sample of whiskey was diluted to 500.0 mL. A 4.004.00 mL aliquot of the diluted sample was removed and the ethanol, C2H5OHC2H5OH , was distilled into 50.00 mL of 0.022500.02250 M K2Cr2O7K2Cr2O7 and oxidized to acetic acid. 3C2H5OH+2Cr2O2−7+16H+⟶4Cr3++3CH3COOH+11H2O3C2H5OH+2Cr2O72−+16H+⟶4Cr3++3CH3COOH+11H2O The excess Cr2O2−7Cr2O72− was then back titrated with 23.523.5 mL of 0.13300.1330 M Fe2+Fe2+ , producing Cr3+Cr3+ and Fe3+Fe3+ . Calculate the weight per volume percent (% w/v) of ethanol in the original whiskey sample.
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.03742 M EDTA solution. The solution is then back titrated with 0.02190 M Zn2 solution at a pH of 5. A volume of 20.48 mL of the Zn2 solution was needed to reach the xylenol orange end point. A 2.000-mL aliquot of the Cu2 and Ni2 solution is fed through an ion-exchange column that retains Ni2 . The Cu2 that passed through the...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.05893 M EDTA solution. The solution is then back titrated with 0.02306 M Zn2 solution at a pH of 5. A volume of 19.89 mL of the Zn2 solution was needed to reach the xylenol orange end point. A 2.000-mL aliquot of the Cu2 and Ni2 solution is fed through an ion-exchange column that retains Ni2 . The Cu2 that passed through the...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.04503 M EDTA solution. The solution is then back titrated with 0.02327 M Zn2 solution at a pH of 5. A volume of 22.80 mL of the Zn2 solution was needed to reach the xylenol orange end point. A 2.000-mL aliquot of the Cu2 and Ni2 solution is fed through an ion-exchange column that retains Ni2 . The Cu2 that passed through the...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.05832 M EDTA solution. The solution is then back titrated with 0.02380 M Zn2 solution at a pH of 5. A volume of 18.49 mL of the Zn2 solution was needed to reach the xylenol orange end point. A 2.000-mL aliquot of the Cu2 and Ni2 solution is fed through an ion-exchange column that retains Ni2 . The Cu2 that passed through the...
What is the concentration of Cu in a solution if a 25.00 mL aliquot reacted with...
What is the concentration of Cu in a solution if a 25.00 mL aliquot reacted with an excess of KI that requires 15.64 mL of 39.94 mM Na2S2O3 solution to titrate the liberated iodine?
A 45.20 mL aliquot from a 0.500 L solution that contains 0.420 g of MnSO 4...
A 45.20 mL aliquot from a 0.500 L solution that contains 0.420 g of MnSO 4 ( MW = 151.00 g/mol) required 35.3 mL of an EDTA solution to reach the end point in a titration. What mass, in milligrams, of CaCO 3 ( MW = 100.09 g/mol) will react with 1.40 mL of the EDTA solution? in mg
A 45.50 mL aliquot from a 0.470 L solution that contains 0.490 g of MnSO4 (Fw...
A 45.50 mL aliquot from a 0.470 L solution that contains 0.490 g of MnSO4 (Fw = 151.00 g/mol) required 40.3 mL of an EDTA solution to reach the end point in a titration. What mass (in milligrams) of CaCO3 (FW = 100.09 g/mol) will react with 1.84 mL of the EDTA solution?
A 20 ml aliquot of malonic acid solution was treated with 10.0 ml of 0.25M Ce4+...
A 20 ml aliquot of malonic acid solution was treated with 10.0 ml of 0.25M Ce4+ leading to the reaction CH2(COOH)2 + 6Ce4+ + 2H2O → HCOOH + 2CO2 + 6Ce3+ + 6H+ After standing for 10 minutes at 60°C, the solution was cooled and the x’ss Ce4+ was titrated with 0.1M Fe2+, requiring 14.4 ml to reach the ferroin end point. Calculate the M of the malonic in the sample.
A 66.0 mL sample of 1.0 M NaOH is mixed with 50.0 mL of 1.0 M...
A 66.0 mL sample of 1.0 M NaOH is mixed with 50.0 mL of 1.0 M H2SO4 in a large Styrofoam coffee cup; the cup is fitted with a lid through which passes a calibrated thermometer. The temperature of each solution before mixing is 23.7 °C. After adding the NaOH solution to the coffee cup, the mixed solutions are stirred until reaction is complete. Assume that the density of the mixed solutions is 1.0 g/mL, that the specific heat of...
ADVERTISEMENT
Subjects
ADVERTISEMENT
Latest Questions
ADVERTISEMENT