Molarity and Molality Formulas & Example Problems

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The colligative nature of the solution is one of the chemicals in class XI SMA. In order to understand this material, you must master the formulas of molarity and molality of a solution well because this chapter has a close relationship with the concentration of a solution.

For those of you who don't know, a solution is a homogeneous mixture of solvents (usually water or H2O) with the dissolved substance. To make it easier for you to understand this material, here we present it from understanding to examples of questions about molarity and molality of a solution.

List of contents

Understanding Molarity

Understanding Molarity

Molarity is a measure of solubility which expresses the number of moles of a solute per volume of solution. The symbol of molarity is the notation M (large) with the molar units being equivalent to moles/liter.

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Molarity Formula

Molarity and Molality Formulas & Example Problems

If earlier was the understanding of molarity, then what about the formula? Well, here is the molarity formula:

Molarity formula

M = molarity of the substance (Molar or mole/L)

n = moles of a substance (moles)

V = volume of solution (ml)

In addition to the formula above, you can also calculate molarity using the following formula:

2 Molarity Formula

M = molarity of the substance (Molar or mole/L)

g = mass of substance (grams)

Mr = relative molecular mass of the substance (grams/mole)

V = volume of solution (ml)

If the solution is mixed, the formula applies:

Solution Mixing Molarity Formula

Va = volume of substance a

Vb = volume of substance b

Ma = molarity of the substance a

Mb = molarity of substance b

If the solution is diluted, then the molarity formula that applies is as follows:

Molarity of substances a and b

Va = volume of substance a

Vb = volume of substance b

Ma = molarity of the substance a

Mb = molarity of substance b

In order to get the mole value, you have to calculate it using the formula below:

5. MOlarity Formula

n = moles of a substance (moles)

g = mass of a substance (grams)

Mr = relative molecule of a substance (grams/mole)

Definition of Molality

Definition of Molality

Similar at first glance, but different molarity you know with molality!

Molality or molal concentration is the number of moles of a solute per one kilogram of solvent. By definition, molality is a quantity that states the number of moles of solute per unit weight of solvent. Molality is denoted by the notation m (small).

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Molality Formula

Molality Formula
Molality Formula

m = molality of a substance (molality)

n = moles of a substance (moles)

p = mass of solvent (grams)

In addition, you can also calculate molality using the following formula:

Molality Formula 2

m = molality of a substance (molality)

Mr = relative molecule of a substance (grams/mole)

p = mass of solvent (grams)

Difference between Molarity and Molality Formulas

Difference between Molarity and Molality Formulas

The terms 'molarity' and 'molality' often make some people confused because of their similar writing and pronunciation. Although both are used to measure the concentration of a solution, there are differences between the two.

Based on the above formula, it can be seen that the difference between molarity and molality lies in the approach. Molarity or molar concentration measures the number of moles of a substance per liter of solution. While molality measures the number of moles of a substance per kilogram of solvent.

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Examples of Molarity Problems and Discussion

Examples of Molarity Problems and Discussion

Now, after knowing the formula for molarity and molality, it would be better if you immediately tried to work on the problem to better understand the material. Come on, just do the following questions!

Problem of molarity

1. If it is known that the mass of the solute is 11 grams, the volume of the solution is 500 mL, and Mr = 22. What is the molarity?

Answer :

Problem Molarity No. 1

So, the molarity of the substance is 1 mol/L.

2. 5 M HCl with a volume of 20 ml was diluted to 0.4 M. So, how much water should be added to the solution?

Answer :

Problem of Molarity No. 2

So, the amount of water that needs to be added is 250 ml.

3. Glucose (C6H12O6) which has a relative mass of 180 is present in 500 mL of a 0.1 M glucose solution. what is the mass of glucose?

Answer :

Molarity Problem No. 3

So, the mass possessed by glucose (C6H12O6) is 9 grams.

4. What is the volume of solvent in H. solution?2SO4 0.2 M whose solute is 0.3 mol?

Answer :

Molarity Problem No. 4

So, the volume of the solvent is 0.15 liters.

5. A total of 18 grams of glucose (C6H12O6) with Mr = 180 is dissolved in water so that the volume becomes 0.5 L. What is the molarity of the solution?

Answer :

Molarity Question No. 5

First of all, calculate the number of moles first

Next, calculate the molarity value.

Molarity Problem No. 5b

So, the molarity of the glucose solution is 0.2 moles per liter.

6. To make 100 mL of a 01, M NaOH solution, how many grams of NaOH (Mr = 40) must be dissolved?

Answer :

Example of molarity question number 6

So, it takes 0.4 grams of the required mass of NaOH.

Molality problem

1. It is known that a solution has a mass of 11 grams of solute, 200 grams of solvent, and Mr = 22. So, what is the molality?

Answer :

Molality problem number 1

Thus, the molality value of 2.5 m is obtained.

2. A total of 250 grams of water is added to 17.1 grams of cane sugar solution which has a Mr = 342. What is the molality of the solution?

Answer :

Molality question number 2

So, the molality of the solution is 0.2 m.

3. What is the molality of a solution of 4 grams of NaOH (if it is known that Ar Na = 23, O = 16, H = 1) dissolved in 250 grams of water?

Answer :

Before calculating molality, we are required to calculate the relative mass first.

Mr = 23 + 16 + 1 = 40 grams/mol

After that, let's calculate the molality.

Molality question number 3

So, the molality contained in the NaOH solution is 0.4 m.

4. What is the molality of a 37% HCl solution if it is known that Ar H = 1 and Ar Cl = 35.5 ?

Answer :

Mr = 1 + 35.5 = 36.5 grams/mol

Mass of HCl = 37% x 100 grams = 37 grams

Mass of solvent = mass of solution – mass of HCl

Mass of solvent = 100 – 37 grams

Mass of solvent = 63 grams or 0.063 kilograms

Molality question number 4

After knowing the moles of HCl, then calculate the molality

Molality question number 4 continued

Thus, it can be seen that the molality of HCl is 16.03 molal.

5. What is the molality of 15 grams of urea CO(NH .?2)2 dissolved in 250 grams of water? (Mr = 60)

Answer :

Calculate the moles of urea first

Molality question number 5

Next, calculate the molality of the solution

Example of Molality Question no 5b

So, the molality of the urea solution is 1 m.

Well, that's the formula for molarity and molality accompanied by the definition and examples of the problem. Easy isn't it? Hopefully through this article you can improve your understanding, yes. Spirit!

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