There are 5 basic types of molarity problems

Molarity or molar concentration uses to describe the number of moles of the substances contained in 1L of solution.Thus,when we caculate the molarity of a solution, it means we calculate the moles per L in a solution.


I will introduce the 5 basic types of molarity problems to help you understand!!!


1.Making a solution with a given concentration


c=n/v , where n= mass(g)×1mol/molar mass(g),so the unit of n is mol.

In this type of question, you will be given the concentration(molarity) of the solution, and calculate the moles(masses) or the volume of the solution. You can according to this equation to calculate the result.


ex1.

What is the [NaCl] in a solution containing 5.12g of NaCl in 250.0 mL of solution?

Moles of NaCl

5.12g×1mol/58.5g =0.0875mol

&

[NaCl]=c=n/V=0.0875mol/0.2500L =0.350M


2. Dilution of a single solution


CDIL=CCONC×VCONC/VDIL

When we dilute a solution,the concentration of the original solution changed. Thus,we should calculate the ratio of two two different solutions, which means we can calulate the ratio of the volumn of the original solution to the volumn of the diluted solution. Then, we multiply the ratio and the original concentration of the solution.


ex.2

A student mixes 100.0 mL of water with 25.0 mL of a sodium chloride solution having an unknown concentration. If the student finds the molarity of the sodium chloride in the diluted solution is 0.0876 M, what is the molarity of the original sodium chloride solution?

The diluted volume is 100.0 mL + 25.0 mL=125.0 mL

Therefore
CCONC=CDIL×VDIL/VCONC=0.0876M×125.0mL/25.0mL=0.438M




3.Mixing two solutions



CDIL(#1)=CCONC(#1) ×VCONC (#1)/VDIL

&

CDIL(#2)=CCONC(#2) ×VCONC(#2)/VDIL
C(total)= CDIL(#1) CDIL(#2)



When we mixing two different solution, we should calculate the concentrations of two different solutions one by one. Fiirst, we calculate the first solution using the previous equation. Then,we use the same equation to calculate the second solution. Finally, we add two concentrations together.


ex.3


If 300.0mL of 0.250M NaCl is added to 500.0mL of 0.100M NaCl, what is the resulting [NaCl] in the mixture?

[NaCl]
DIL(#1)=[NaCl]CONC(#1)×VCONC(#1)/VDIL=0.250M×300.0mL/800.0mL=0.09375M
[NaCl](#2)=0.100M×500.0mL/800.0mL=0.06250M
[NaCl](total)=[NaCl]DIL(#1)+[NaCl]DIL(#2)=0.09375+0.06250=0.156M


4.Converting a density to a molarity and converting a molarity to a density


c= d (g)/(L)
× 1mol/molar mass(g) & d=c (mol)/(L) × molar mass(g)/1mol


Allthough the questions are different, the idea is the same.When we are asked to calculate the molarity of the solution, it means we should get the unit of mol/L. In the question, we know the density of the solution,which means we can get the unit of g/L. In order to get the unit of mol/L, we should use the molar mass(mol/g).Thus, the g will be cancealed, and we can get the answer that we want. For the question of converting a molarity to a density, it uses the same idea.


ex.4

What is the molarity of pure sulphuric acid, H2SO4, having a density of 1.839 g/mL?

[H
2SO4]=1.839g/0.001L×1mol/98.1g=18.7M




5.Making a dilute solution from a concentrated solution



CCONC×VCONC=CDIL×VDIL



The equation shows that the moles of the concentrated solution and the moles of the diluted solution are the same, and it uses the same idea as the second types. Thus, when we want to calculate one of the factors above, we can use this equation.

ex.5

What volume of 6.00M HCl is used in making up 2.00L of 0.125M HCl?

V
CONC=CDIL×VDIL/CCONC=0.125M×2.00L/6.00M=0.0417L