The concentration in moles per dm^{3} can be used to determine the number of moles present in any (other) volume of a solution, V cm^{3}. As in, all mole calculations, starts by determining the amount in moles of the solution, for which it is the molar concentration and has measured the volume used.

A mixture is formed when various elements or compounds just mix together without any chemical reaction taking place. Unlike a compound, a mixture has a variable composition with its constituents showing their individual properties. When salt is added to a beaker of water, salt readily dissolves in water without any chemical reaction taking place. Here, salt is a compound with the elements sodium and chlorine and water is another compound comprising the elements hydrogen and oxygen. Thus, we can say that all matter can be divided into elements, compounds and mixtures

The balanced chemical equation leads to the mole ratio between the acid and the base which in turn leads to the amount in moles of the base. Finally molar concentration of the base is calculated from the known volume and the amount in moles.

Related Calculators | |

Molar Concentration Calculator | concentration dilution calculator |

ph concentration calculator | Calculate Molarity |

Solved problems based on molar concentration are given below. ### Solved Examples

**Question 1: **Calculate the molar concentration of NaOH for the reaction between HCl and NaOH.

** Solution: **

The balanced chemical equation is

HCl + NaOH $\rightarrow$ NaCl + H_{2}O

For an acid

n(HCl) = $\frac{35.0}{1000}$dm^{3} $\times$ 0.250 mol dm^{-3} = 8.75 $\times$ 10^{-3}mol

Therefore the mole ratio NaOH:HCl = 1:1. The amount in moles of NaOH present is 8.75 $\times$ 10^{-3}mol.

Now we use the equation.

but this time for a base. First convert the volume of aqueous NaOH used to dm^{-3}.

25.0cm^{3} = $\frac{25.0}{1000}$dm^{3} = 25.0 $\times$ 10^{-3}dm^{3}

Therefore molar concentration of NaOH = $\frac{8.75 \times 10^{-3}mol}{25.0 \times 10^{-3}dm^{3}}$

Molar concentration of NaOH = 0.350 mol dm^{-3}

**Question 2: **The concentration of Ca(HCO_{3})_{2} is 0.74 gmol/L. Convert this concentration to geq/L.

** Solution: **

[C]_{eq} = $\frac{[C][MM]}{eq.mass}$

[C] = 0.74 gmol/ L

MM = 40.1(2) + 2{1+12+3(16)} = 202.2

No. of reference species = 2

Therefore eq.mass = $\frac{Ca(HCO_{3})_{2}}{2}$ = $\frac{202.2}{2}$

[C]_{eq} = $\frac{0.74[202.2]}{202.2/2}$ = 1.48 geq/L

The balanced chemical equation is

HCl + NaOH $\rightarrow$ NaCl + H

For an acid

n(HCl) = $\frac{35.0}{1000}$dm

Therefore the mole ratio NaOH:HCl = 1:1. The amount in moles of NaOH present is 8.75 $\times$ 10

Now we use the equation.

but this time for a base. First convert the volume of aqueous NaOH used to dm

25.0cm

Therefore molar concentration of NaOH = $\frac{8.75 \times 10^{-3}mol}{25.0 \times 10^{-3}dm^{3}}$

Molar concentration of NaOH = 0.350 mol dm

[C]

[C] = 0.74 gmol/ L

MM = 40.1(2) + 2{1+12+3(16)} = 202.2

No. of reference species = 2

Therefore eq.mass = $\frac{Ca(HCO_{3})_{2}}{2}$ = $\frac{202.2}{2}$

[C]