Chemistry Quiz On Molarity! Trivia

1. What is the molarity of a solution that contains 3 moles of Lithium Chloride in 2 Liters of the solution?

.66 M LiCl solution

1.5 M LiCl solution

6 M LiCl solution

2 M LiCl solution

Molarity: Divide moles by number of liters (3 moles / 2 Liters)

Explanation

Molarity: Divide moles by number of liters (3 moles / 2 Liters)

2. What is the molarity of a 300 mL solution containing 66.588 g of Calcium chloride?

0.002 M solution

2.0 M solution.

0.02 molar solution

20.0 molar solution

Calcium chloride is CaCl2. So the molar mass is 110.98. 300 ml of solution = .3 Liters

Explanation

Calcium chloride is CaCl2. So the molar mass is 110.98. 300 ml of solution = .3 Liters

3. What is the molarity of a solution containing 3.646 g of HCl (hydrochloric acid) in a 400 ml Solution?

1 M HCl solution

1.25 HCl solution

.25 M HCl Solution

0.025 M HCl Solution

Change grams to moles. (Divide by molar mass). Change mL to Liters (400 mL = .4 L)

Explanation

Change grams to moles. (Divide by molar mass). Change mL to Liters (400 mL = .4 L)

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

1.6 M LiF solution

4 M LiF solution

1.2 M LiF solution

.4 Molar solution

The molarity of a solution is calculated by dividing the moles of solute by the volume of solution in liters. In this case, we need to first convert the given mass of LiF to moles using its molar mass. Then, we divide the moles of LiF by the volume of the solution in liters (which is 0.250 L since it is given in milliliters). The resulting value is 1.6 M LiF solution.

Explanation

The molarity of a solution is calculated by dividing the moles of solute by the volume of solution in liters. In this case, we need to first convert the given mass of LiF to moles using its molar mass. Then, we divide the moles of LiF by the volume of the solution in liters (which is 0.250 L since it is given in milliliters). The resulting value is 1.6 M LiF solution.

5. How much CaCl₂ and water is needed to make a .27 M solution?

29.96 g of Calcium chloride with enough distilled water to make 1 Liter.

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

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

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

First, find the molar mass of CaCl2. (One calcium and TWO chlorides). Multiply the molar mass by .27. Add water to the 1-liter mark.

Explanation

First, find the molar mass of CaCl2. (One calcium and TWO chlorides). Multiply the molar mass by .27. Add water to the 1-liter mark.

6. 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?

Add 8.3 mL of 12 M HCl to a graduated cylinder and add water to the 1 L mark.

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

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

(molarity 1) (Volume 1) = (molarity 2) (volume 2) (12 M HCl) (volume 1) = (.1 M HCl) (100 mL)

Explanation

(molarity 1) (Volume 1) = (molarity 2) (volume 2) (12 M HCl) (volume 1) = (.1 M HCl) (100 mL)

7. How much LiF and water are needed to make a .7 M solution?

Add 18.16 grams of LiF and 1 Liter of distilled water.

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

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

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

Find the molar mass of LiF (look on the periodic table). since a 0.7 M solution is required, multiply the molar mass by 0.7. Add water to the 1-liter mark.

Explanation

Find the molar mass of LiF (look on the periodic table). since a 0.7 M solution is required, multiply the molar mass by 0.7. Add water to the 1-liter mark.

8. 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.

To prepare a 0.3 M solution of NaCl, you need to calculate the molar mass of NaCl by adding the atomic masses of sodium (22.99 g/mol) and chlorine (35.45 g/mol). Then, multiply the molar mass by 0.3 to determine the amount of NaCl required. Place this amount of NaCl in a volumetric flask and add distilled water to the 1 L mark. Stir the solution to ensure complete dissolution of NaCl.

Explanation

To prepare a 0.3 M solution of NaCl, you need to calculate the molar mass of NaCl by adding the atomic masses of sodium (22.99 g/mol) and chlorine (35.45 g/mol). Then, multiply the molar mass by 0.3 to determine the amount of NaCl required. Place this amount of NaCl in a volumetric flask and add distilled water to the 1 L mark. Stir the solution to ensure complete dissolution of NaCl.

9. 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.

300 ml of a 6 M Solution

8.33 ml of a .5 M Solution

8.33 ml of a 3 M HCl solution

8.33 ml of 6 M HCl solution

The questions ask for the amount of fluid from a 6 M solution. So, the answer must be either A or D. To solve the problem: 0.5 M x (100 ml) divided by (6 M)

Explanation

The questions ask for the amount of fluid from a 6 M solution. So, the answer must be either A or D. To solve the problem: 0.5 M x (100 ml) divided by (6 M)

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?

Add 50.0 mL of 12 M HCl to a graduated cylinder and add water to the 1 L mark.

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

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

(molarity 1) (Volume 1) = (molarity 2) (volume 2) (12 M HCl) (volume 1) = (3.0 M HCl) (200 mL)

Explanation

(molarity 1) (Volume 1) = (molarity 2) (volume 2) (12 M HCl) (volume 1) = (3.0 M HCl) (200 mL)