10 Questions
| Attempts: 605

Questions and Answers

- 1.How do you prepare a .3
*M*solution of NaCl? The molar mass of NaCl is 22.99 + 35.45. Since a .3 M solution is required, multiply the molar mass by .3. Water is added to the 1-liter mark. REMEMBER: Molarity is Moles per Liter solution. The liter solution INCLUDES the water AND the salt -- salt has volume -- so it will be less than 1 Liter of water.- A.
Find the molar mass of NaCl. Add that amount of NaCl to a volumetric flask and add distilled water to the 1 L mark. Stir.

- B.
Find the molar mass of NaCl. Add that amount of NaCl to a volumetric flask and add 1 Liter of distilled water. Stir.

- C.
Look up the molar mass of sodium and chlorine on the periodic table. Multiply the molar mass by .3. Place this amount in a volumetric flask and add distilled water to the 1 L mark. Stir.

- D.
Look up the molar mass of sodium and chlorine on the periodic table. Multiply the molar mass by .3. Place this amount in a volumetric flask and add 1 liter of distilled water. Stir.

- 2.How much CaCl
_{2 }and water is needed to make a .27*M*solution?- A.
29.96 g of Calcium chloride with enough distilled water to make 1 Liter.

- B.
110.98 g of Calcium chloride with enough water to equal 1 liter of solution.

- C.
29.96 g of Calcium chloride with 1 liter of distilled water.

- D.
110.98 g of Calcium chloride with 1 liter of distilled water.

- 3.How much LiF and water are needed to make a .7 M solution?
- A.
Add 18.16 grams of LiF and 1 Liter of distilled water.

- B.
Add 25.94 grams of LiF and 1 Liter of distilled water.

- C.
Add 25.94 g of LiF to a volumetric flask with enough distilled water to reach the 1 Liter mark.

- D.
Add 18.16 grams of LiF with enough distilled water to make 1 Liter of solution.

- 4.What is the molarity of a solution that contains 10.376 grams of LiF in a 250 ml solution?
- A.
1.6 M LiF solution

- B.
4 M LiF solution

- C.
1.2 M LiF solution

- D.
.4 Molar solution

- 5.What is the molarity of a 300 mL solution containing 66.588 g of Calcium chloride?
- A.
0.002 M solution

- B.
2.0 M solution.

- C.
0.02 molar solution

- D.
20.0 molar solution

- 6.What is the molarity of a solution containing 3.646 g of HCl (hydrochloric acid) in a 400 ml Solution?
- A.
1 M HCl solution

- B.
1.25 HCl solution

- C.
.25 M HCl Solution

- D.
0.025 M HCl Solution

- 7.What is the molarity of a solution that contains 3 moles of Lithium Chloride in 2 Liters of the solution?
- A.
.66 M LiCl solution

- B.
1.5 M LiCl solution

- C.
6 M LiCl solution

- D.
2 M LiCl solution

- 8.How is a 100 mL solution of .5 M HCl prepared from a stock of 6 M HCl solution? (Hint: This is dilution problem : (molarity of solution 1) (volume of solution 1) = (molarity of solution 2) (volume of solution 2). Dilution problems: You are solving for V1. Set the equation: (6 M HCl) (Volume) = (.5 M HCl) (100 ml).
*IF v2 is in ml, then v1 will also be in mL.*- A.
300 ml of a 6 M Solution

- B.
8.33 ml of a .5 M Solution

- C.
8.33 ml of a 3 M HCl solution

- D.
8.33 ml of 6 M HCl solution

- 9.A teacher has a 12 M stock solution of HCl. Her students need 100 mL of .1 M solution to complete their lab. How is it prepared?
- A.
Add 8.3 mL of 12 M HCl to a graduated cylinder and add water to the 1 L mark.

- B.
Add 83 ml of 12 M HCl solution to volumetric flask and add water to the 1 Liter mark.

- C.
Take 0.83 mL of 12 M HCl and add it to a graduated cylinder; add distilled water to the 100 mL mark to get a .1 M solution.

- D.
Add 10 mL of 12 M HCl to a graduated cylinder and add water to the 100 ml mark.

- 10.A teacher has a 12 M stock solution of HCl. Her students need 200 mL of 3.0 M solution to complete their lab. How is it prepared?
- A.
Add 50.0 mL of 12 M HCl to a graduated cylinder and add water to the 1 L mark.

- B.
Add 83 ml of 12 M HCl solution to volumetric flask and add water to the 1 Liter mark.

- C.
Take 5.0 of 12 M HCl and add it to a graduated cylinder; add distilled water to the 100 mL mark to get a .1 M solution.

- D.
Add 10 mL of 12 M HCl to a graduated cylinder and add water to the 100 ml mark.

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