# Molarity - Concentration of Solutions

To study solution stoichiometry, we must know how much of the reactants are present in a solution and also how to control the amounts of reactants used to bring about a reaction in aqueous solution.

####
**The
concentration of a solution**

** The
concentration of a solution is the amount of solute present in a given amount of
solvent, or a given amount of solution. (For this discussion, we will assume
the solute is a liquid or a solid and the solvent is a liquid.)

** The
concentration of a solution can be expressed in many different ways. Here we will
consider one of the most commonly used units in chemistry, molarity (M), or molar
concentration.

####
**Molarity
(M)**

** Molarity is The
number of moles of solute per liter of solution.

where (n)
denotes the number of moles of solute and (V) is the volume of the solution in
liters.

** A 1.46 molar
glucose (C

_{6}H_{12}O_{6}) solution, written as 1.46 M (C_{6}H_{12}O_{6}), contains 1.46 moles of the solute (C_{6}H_{12}O_{6}) in 1 L of the solution. Of course, we do not always work with solution volumes of 1 L. Thus, a 500-mL solution containing 0.730 mole of (C_{6}H_{12}O_{6}) also has a concentration of 1.46 M:
** Note that
concentration, like density, is an intensive property, so its value does not depend
on how much of the solution is present.

** It is
important to keep in mind that molarity refers only to the amount of solute
originally dissolved in water and does not take into account any subsequent
processes, such as the dissociation of a salt or the ionization of an acid.
Consider what happens when a sample of potassium chloride (KCl) is dissolved in
enough water to make a 1 M solution:

** Because KCl
is a strong electrolyte, it undergoes complete dissociation in solution. Thus,
a 1 M KCl solution contains 1 mole of K

^{+}ions and 1 mole of Cl^{-}ions, and no KCl units are present. The concentrations of the ions can be expressed as [K^{+}] = 1 M and [Cl^{- }] = 1 M, where the square brackets [ ] indicate that the concentration is expressed in molarity. Similarly, in a 1 M barium nitrate [Ba(NO_{3})_{2}] solution:
we have [Ba

^{2+}] = 1 M and [NO_{3}^{- }] = 2 M and no Ba(NO_{3})_{2}units at all.####
**Preparing
a solution of known molarity **

The procedure
for preparing a solution of known molarity is as follows

**(1)**The solute is accurately weighed and transferred to a volumetric flask through a funnel.

**(2)**Water is added to the flask, which is carefully swirled to dissolve the solid.

**(3)**After all the solid has dissolved, more water is added slowly to bring the level of solution exactly to the volume mark.

Knowing the
volume of the solution in the flask and the quantity of compound (the number of
moles) dissolved, we can calculate the molarity of the solution using Equation
(1).

Note that this
procedure does not require knowing the amount of water added, as long as the volume
of the final solution is known.

####
**Solved
Problems**

**Example (1): How many grams of potassium dichromate (K**

_{2}Cr_{2}O_{7}) are required to prepare a 250 mL solution whose concentration is 2.16 M ?**Strategy**

How many moles of K

_{2}Cr_{2}O_{7}does a 1-L (or 1000 mL) 2.16 M K_{2}Cr_{2}O_{7}solution contain? A 250-mL solution? How would you convert moles to grams?**Solution**

The first step is to determine the number of
moles of K

_{2}Cr_{2}O_{7}in 250 mL or 0.250 L of a 2.16 M solution. Rearranging Equation (1) gives:
The molar mass
of K

_{2}Cr_{2}O_{7}is 294.2 g, so we write:**Check**

As a ball-park estimate, the mass should be
given by:

[molarity
(mol/L) × volume (L) × molar mass (g/mol)] =

[2 mol/L × 0.25
L × 300 g/mol] = 150 g.

So the answer
is reasonable.

**Example (2): In a biochemical assay, a chemist needs to add 3.81 g of glucose to a reaction mixture. Calculate the volume in milliliters of a 2.53 M glucose solution she should use for the addition.**

**Strategy**

We must first
determine the number of moles contained in 3.81 g of glucose and then use
Equation (2) to calculate the volume.

**Solution**

From the molar
mass of glucose, we write:

Next, we calculate
the volume of the solution that contains 2.114 × 10

^{-2}mole of the solute. Rearranging Equation (2) gives:**Check**

One liter of
the solution contains 2.53 moles of C

_{6}H_{12}O_{6}. Therefore, the number of moles in 8.36 mL or 8.36 10^{-3}L is (2.53 mol × 8.36 × 10^{-3}) or 2.12 × 10^{-2}mol. The small difference is due to the different ways of rounding off.

*Reference: Chemistry*

*/ Raymond Chang ,*

*Williams College*

*/*

*(10th edition).*

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