**Molar concentration (M) = $\frac{Moles\ of\ solute(n)}{Volume\ of\ solution\ (V in liter)}$**

Where Mole of solute defined as mass of solute in grams divided by molar mass of solute.

**Moles of solute (n) = $\frac{Mass\ of\ solute\ (w in grams)}{Molar\ mass\ of\ solute (W)}$**

So if in any problem, mass (gram) of solute (w) is given then first we have to divide it by molar mass (molecular weight) of solute (W) in order to find the moles of solute (n) present in the solution, and again divide moles of solute (n) by volume (liter) of solution (V) in order to find molarity (M) or molar mass of solution.

So we can also write this molar concentration formula (molarity) like this,

**Molar concentration (M) = $\frac{Mass\ of\ solute\ (w in grams)}{Molar\ mass\ of\ solute (W)} \times [Volume of solution (V in liter)]$**

So on the basis of above given molar concentration formula we can say that if one mole of any solute dissolved in one liter of solution then that solution is known as molar (1M) solution.

**Millimolar (mM) = 10 ^{-3} Molar (M)**

**Micromolar (μM) = 10 ^{-6} Molar (M)**

Related Calculators | |

Molar Concentration Calculator | concentration dilution calculator |

ph concentration calculator | Calculate Molarity |

As in all mole calculations this starts by determining the amount in moles of the solution for which we know the molar concentration and have measured the volume used. The molar concentration is calculated from the known volume and the amount in moles.

Example problem is given below.### Solved Example

**Question: **Find out the molar concentration of acetic acid (CH_{3}COOH) solution when 1.24 grams of acetic acid (CH_{3}COOH) dissolved in 155.0 ml solution?

** Solution: **

Example problem is given below.

Given

mass of acetic acid (w) = 1.24 g

mass of acetic acid (w) = 1.24 g

volume of acetic acid (V) = 155.0 ml

= (155.0 / 1000) L {Because 1 Liter = 1000 ml}

= 0.155 L

Molar mass of acetic acid (W) = atomic mass of C atom + atomic mass of 2 O atoms + atomic mass of 4 H atoms

= (12.0 + 2 * 16.00 + 4 * 1.0) g / mol

= 48.0 g / mol

= 48.0 g / mol

And now by the help of molar concentration (molarity) formula we can determine molar concentration of acetic acid,

Molar concentration of acetic acid (M) = mass of acetic acid (w) / molar mass of acetic acid (W) * volume of acetic acid (V)

= 1.24 g / (48.0 g / mol * 0.155 L)

= 0.167 M

So when 1.24 grams of acetic acid (CH_{3}COOH) dissolved in 155.0 ml solution, molar concentration of this solution is 0.167 M.