Chemistry Quiz On Molarity! Trivia

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Do you know what the term molarity means? M denotes molarity, and it is defined as the number of moles of solute present in one liter of solution. To obtain molarity, you will need to divide the moles of the solute by liters of solution. It is most expressed in units of moles. This chemistry quiz on molarity is top-notch. It would be best if you simply took this incredible quiz.

• 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.

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.
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.

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• 2.

How much CaCl2   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.

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

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• 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.

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

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• 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

A. 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.

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• 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

B. 2.0 M solution.
Explanation
Calcium chloride is CaCl2. So the molar mass is 110.98. 300 ml of solution = .3 Liters

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• 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

C. .25 M HCl Solution
Explanation
Change grams to moles. (Divide by molar mass). Change mL to Liters (400 mL = .4 L)

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• 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

B. 1.5 M LiCl solution
Explanation
Molarity: Divide moles by number of liters (3 moles / 2 Liters)

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• 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

D. 8.33 ml of 6 M HCl solution
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)

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• 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.

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

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• 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.

A. Add 50.0 mL of 12 M HCl to a graduated cylinder and add water to the 1 L mark.
Explanation
(molarity 1) (Volume 1) = (molarity 2) (volume 2) (12 M HCl) (volume 1) = (3.0 M HCl) (200 mL)

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• Current Version
• Mar 22, 2023
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• Apr 26, 2012
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