# Molarity Quick Review and Practice Questions

- Posted by Brian Stocker MA
- Date Published April 7, 2014
- Date modified January 12, 2019
- Comments 11 comments

## Molarity

Molarity is the measure of the concentration of a substance in a solution, given in terms of the amount of substance per unit volume of the solution. Molarity questions are on the HESI and the NLN PAX

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### How to Solve Molarity Questions

Molarity is also called, amount-of-substance concentration, amount concentration, substance concentration, or simply concentration.

The Molarity of a solution simply means the amount of moles contained in every liter of a solution. To better understand the concept of molarity of a solution it is necessary to first understand some related terms.

* **A solute** – is the substance that is being dissolved such as sugar or mercury.

* **A solvent** – refers to the liquid that the solute is being dissolved in.

* **A solution** – refers to the mixture of the solvent and the solute so that solution equals solvent plus solute.

The Molarity of the solution is thus a measurement of the molar concentration of the solute in the solution. The molarity of a solution is measured in moles of solute per liter of solution, or mol/liter. For example, if the molarity of a mercury solution is 1M, it simply means that there is 1 mole of sugar contained in every 1 liter of the solution.

The formula for molarity is = moles of solute/total liters of solution

### Here is a typical question:

**If 2 moles of salt is dissolved to form 1 liter of solution, calculate the molarity of the solution.**

a. 1 M solution

b. 1.5 M solution

c. 2 M solution

d. 2.5 M solution

The formula for calculating molarity when the moles of the solute and liters of the solution are given is = moles of solute/ liters of solution.

Moles of Solute = 2 moles of sugar

Solution liters = 1 liters

The molarity of the solution = 2 moles of solvent/1 liters of solution = 2 M solution.

### Practice Questions

**1. Calculate the molarity of a sugar solution if 4 liters of the solution contains 8 moles of sugar?**

a. 0.5 M

b. 8 M

c. 2 M

d. 80 M

**2. What is the molarity of a solution containing 5 moles of solute in 250 milliliters of solution?**

a. 20 M

b. 15 M

c. 0.104 M

d. 1.25 M

**3. ****How many moles of N _{A}OH are needed to dissolve in water to make 4 liters of a 2.0 M solution?**

a. 0.50 M

b. 2 M

c. 8 mol

d. 0.5 mol

**4. How many moles of Na are needed to make 4.5 liters of a 1.5 M Na solution?**

a. 6.75 mol

b. 0.33 M

c. 0.33 mol

d. 3 M

**5. Molarity of a solution can be defined as the:**

a. atomic mass of an element

b. moles of solution per liter of solute

c. moles of solute per liter of solution

d. mass of solvent per liter of solution

**6. To calculate the Molarity of a solution when the solute is given in grams and the volume of the solution is given in milliliters, you must first:**

a. Convert grams to moles, but leave the volume of solution in milliliters

b. Convert volume of solution in milliliters to liters, but leave grams to moles

c. Convert grams to moles, and convert volume of solution in milliliters to liters

d. None of the above

**7. The molarity of an aqueous solution of CaCl is defined as the**

a. moles of CaCl per milliliter of solution

b. grams of CaCl per liter of water

c. grams of CaCl per milliliter of solution

d. moles of CaCl per liter of solution

## Get the Book Students are Talking About!

### Answer Key

**1. C**

Molarity = moles of solute/liters of solution = 8/4 = 2

**2. A**

First convert 250 ml to liters, 250/1000 = 0.25 then calculate molarity = 5 moles/ 0.25 liters = 20 M

**3. C**

A solution with molarity 2 requires 2 M of N_{A}OH per liter.

So, 4 X 2 = 8 M

**4. A
** A solution of molarity 1.5 M, requires 1.5 mol of Na to every litre of solvent.

1.5 mol of Na into 1L renders 1L of 1.5M solution

Therefore, multiply the molarity of the desired solution by the end volume required:

4.5L requires 6.75 mol of Na, as 1.5(M)*4.5(L)=6.75 (mol).

**5. C**

**6. C**

**7. D**

**Written by:**Brian Stocker MA, Complete Test Preparation Inc.

## 11 Comments

It was very easy . I want some difficult questions ..

It was good, even though they were easy, it’s great to have problems

It is very easy

in 3 rd question it will be NaOH but it is NAOH

thanks!

yes.you are correct

thanks HAPPY NEW YEAR

It’s very eazy dont know what I waz stressing about

it was easy . Please upload some difficult and tricky questions.

All questions r easy …I want more questions

thanks.now i know basic molarity issues.do you mind bringing a little bit complex ones?