Free Experimental Techniques in Chemistry MCQs with Answers

11. During the process of crystallization, the hot saturated solution:

• A. Is cooled very slowly to get large-size pores in it
• B. Is cooled at a moderated rate to get medium-sized crystals
• C. Is evaporated to get the crystals of the product
• D. Is mixed with an immiscible liquid to get the pure crystals of the product

Explanation: The hot filtered solution is cooled at a moderate rate so that medium sized crystals are formed. Slow cooling yields bigger crystals which are likely to include considerable amount of solvent carrying impurities with it and complicating the drying process.

12. Solvent extraction method is particularly useful technique for separation when the product to be separated is:

• A. non-volatile or thermally unstable
• B. non-volatile or thermally stable
• C. volatile or thermally unstable
• D. volatile or thermally stable

Explanation: Solvent extraction is particularly useful when dealing with volatile or thermally unstable products because it allows for the separation of components without subjecting them to high temperatures. This method relies on the differential solubility of substances in a chosen solvent. For volatile or thermally unstable compounds, traditional separation methods like distillation can be impractical since they involve heating. Solvent extraction operates at lower temperatures, reducing the risk of degrading or losing the desired product. It's especially effective for extracting delicate compounds from mixtures without compromising thei

13. The comparative rates at which the solutes move in paper chromatography, depend on:

• A. the size of paper
• B. Retention Factor
• C. temperature of the experiment
• D. size of the chromatographic tank used

Explanation: The comparative rate at which the solutes move in paper chromatography mainly depends on the retention factor, which is expressed as the ratio of the distance travelled by the solute to the total distance travelled by the solvent (i.e., solvent front).

14. In the reaction setup shown above, which of the following are true? I. This setup can be used to prepare a soluble gas by water displacement. II. This setup involves a decomposition reaction if the substance heated is potassium chlorate. III. This setup can be used to prepare an insoluble gas by water displacement.

• A. I only
• B. II only
• C. I and III
• D. II and III
• E. I, II, and III

Explanation: This setup can be used to prepare an insoluble gas but not a soluble one. If the substance is potassium chlorate, it does decompose into potassium chloride and oxygen.

15. In this laboratory setup for distillation, where does the vaporization take place?

• A. Around the thermometer
• B. In the condenser
• C. In the circulating water
• D. In the heated flask
• E. In the distillate

Explanation: Option d is correct because in a laboratory setup for distillation, vaporization takes place in the heated flask. This is where the liquid is heated, leading to the formation of vapor or gaseous form of the substance.

16. If the liquid being distilled contains dissolved magnesium chloride, where will it be found after distillation is completed?

• A. Around the thermometer
• B. In the condenser
• C. In the circulating water
• D. In the heated flask
• E. In the distillate

Explanation: The magnesium chloride will be left behind in the heated flask since it is not volatile as the liquid boils off.

17. If the flame used to heat a flask is an orange color and blackens the bottom of the flask, what correction should you make to solve this problem?

• A. Move the flask farther from the flame.
• B. Move the flask closer to the flame.
• C. Allow less air into the collar of the burner.
• D. Allow more air into the collar of the burner.
• E. The problem is in the supply of the gas, and you cannot fix it.

Explanation: One of the basic adjustments to a burner is to assure enough air is mixing with the gas to form a blue cone-shaped flame. With insufficient air, carbon deposits will form on the flask and the flame will be orange.

18. In the above titration setup, if you introduce 15 mL of the NaOH with an unknown molarity into the flask and then add 5 drops of phenolphthalein indicator, what will you observe?

• A. A pinkish color will appear throughout the solution.
• B. A blue color will appear throughout the solution.
• C. There will be a temporary pinkish color that will dissipate.
• D. There will be a temporary blue color that will dissipate.
• E. There will not be a color change.

Explanation: Phenolphthalein indicator is colorless below a pH of 8.3 but is red to pink in basic solutions above this pH. In this NaOH solution, it will be red to pink.

19. If an HCl solution has a concentration of 0.1 M and it requires 30 mL to reach the end point, and during the titration, 15 mL of NaOH solution is used, what is the molarity of the NaOH solution?

• A. 0.1 M
• B. 0.2 M
• C. 0.3 M
• D. 1M
• E. 2 M

Explanation: Calculate the molarity by using the formula: M acid X V acid = M base X V base 0.1 M X 30 mL = M base x 15 mL M base =0.2 M Notice that as long as the units of volume are the same, they cancel out of the equation.

20. When is the end point reached and the volume of the HCl recorded in this reaction? I. When the color first disappears and returns in the flask. II. When equal amounts of HCl and NaOH are in the flask. III. When the color disappears and does not return in the flask.

• A. I only
• B. III only
• C. I and III
• D. II and III
• E. I, II, and III

Explanation: The end point is reached when the color of the indicator disappears and does not return. The color will first disappear temporarily before the end point but finally will not return.