# Solution making

What's the equation you use to make a solution from a solid? For example, I have copper sulfate in a solid powder but need it as a solution.

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Publication Date: 02 April 2016
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## Answer by ginny.r.ward on question Solution making

In brief

A commonly used expression for the concentration of a solute in solution is molar concentration or molarity. Molarity is defined as the number of moles of solute dissolved in one litre of solution, and has the unit mol/L, denoted by upper case M. When preparing a solution where the concentration is expressed as a molarity, the formula to use is

m = c x V x M

where:

m = mass of solute (in grams)

c = concentration of solute (in moles per litre)

V = volume (in litres)

M = molecular weight (or molar mass) of solute (in grams)

The molecular weight can be found on the bottle label, or in a data book or safety data sheet (SDS), or by adding together the atomic weights of all of the atoms which appear in the chemical formula of the substance. If you are using a hydrated salt, the water(s) of hydration must be included in the calculation of the molecular weight.

Example: Preparation of 1 litre of a 0.5 M copper (II) sulfate solution

Starting with copper (II) sulfate pentahydrate, (CuSO4.5H2O), we have

Molecular weight = M = 63.55 + 32.06 + (4x15.99) + 5x((2x1.008) + 15.99) = 249.68 g

Concentration = c = 0.5 M

Volume = V = 1 L

The quantity of the solid copper sulfate pentahydrate required is therefore

m = c x V x M

= 0.5 x 1 x 249.68

= 124.84 g

The copper sulfate should be added to about half to two-thirds of the final solution volume (about 500–650 mL) of distilled water in a large beaker, and stirred until dissolved. Gentle heating will speed up the dissolution of the salt. The solution can then be transferred to a measuring cylinder or volumetric flask and distilled water added to make up the volume to 1 litre.

The solution can also be prepared directly in a volumetric flask. The final volume should be measured with the solution and vessel at room temperature, as this is the temperature at which volumetric glassware is calibrated. If heating the solution, a beaker or conical flask should be used rather than a volumetric flask, as heating volumetric glassware may affect the calibration.

Solutions of copper sulfate can degrade due to hydrolysis of the copper (II) ion with the formation of a precipitate of copper (II) hydroxide (Cu(OH)2). The presence of this precipitate is indicated by a cloudiness in the solution. The copper sulfate solution can be stabilised by addition of a 0.1 M sulfuric acid solution in small aliquots (1–5 mL) until the solution becomes clear.

Example: Preparation of 250 mL of a 0.2 M solution of sodium carbonate, starting with the anhydrous salt (Na2CO3). Before performing the calculation, we convert the volume, 250 mL, to litres, i.e. 0.25 L.

Molecular weight = M = 105.99 g

Concentration = c = 0.2 M

Volume = V = 0.25 L

The quantity of the solid sodium carbonate required is therefore:

m = c x V x M
= 0.2 x 0.25 x 105.99
= 5.30 g

Example: As in the previous example, but with the monohydrate, i.e. 250 mL of a 0.2 M solution of sodium carbonate, starting with sodium carbonate monohydrate (Na2CO3.H2O).

Molecular weight = M = 124.00 g

Concentration = c = 0.2 M

Volume = V = 0.25 L

The quantity of the solid sodium carbonate monohydrate required is therefore:

m = c x V x M
= 0.2 x 0.25 x 124.00
= 6.20 g

The mole

The mole is a unit used to describe the amount of a chemical species. It can be used to describe the amount of atoms, molecules, ions, electrons, etc. The number of the species in one mole is 6.022 x 1023 (Avogadro's number). The abbreviation of mole or moles is mol.

The use of the term mole in chemistry is analogous to how the word dozen is used in everyday language.  For example, one dozen apples is 12 apples, while one mole of apples would be 6.022 x 1023 apples.

The weight in grams of one mole of a substance is the molecular weight, or molar mass, of that substance. To determine the number of moles, n, in a given quantity of a substance, divide the given quantity of the substance by the molecular weight:

n = (mass of substance (g))/(molecular weight (g))  mol

or

n = m/M

Molarity

The unit for molar concentration is molarity, which has the symbol M, and the dimensions, moles per litre.

molar concentration = c = (number of moles)/(volume)   mol/L

or

c = n/V

The symbol M is pronounced ‘molar’; for example, ‘1 M’ is pronounced ‘one molar’, and ‘a 2 M solution’ is pronounced ‘a two molar solution’.

Note that the equation c = n/V can be rearranged if you want to find the volume of a solution of known concentration, which will give you a certain number of moles of the solute.

V = n/c

Or, if you want to find the number of moles of solute in a certain volume of a solution of known concentration

n = c x V

Percentage concentration

The concentration of a solution can also be expressed as a percentage concentration, usually either as %w/v or %v/v and also sometimes as %w/w.

Where the solute is a solid, the percentage concentration would usually be prepared as the percent weight per volume, or %w/v:

%w/v = (mass of solute (g))/(volume of solution (mL))  x 100%

Another way of expressing this is:

%w/v = mass of solute (g) in 100 mL of solution.

For example, a 2% w/v solution of sodium chloride would be prepared from 2 g of sodium chloride dissolved in water and made up to a volume of 100 mL.

Where the solute is a liquid, the percentage concentration can be prepared as the percent volume per volume, or %v/v:

%v/v = (volume of solute)/(volume of solution)   x 100%

Alternatively,

%v/v = volume of solute (mL) in 100 mL of solution.

For example, a 5% v/v aqueous solution of ethanol would be prepared by taking 5 mL of pure ethanol and diluting this with water to a volume of 100 mL.

Weight percent is often used in aqueous commercial preparations, for example, in concentrated solutions of acids. A weight percent concentration has the advantage that the solution can be prepared independently of temperature considerations.

%w/w = (mass of solute)/(mass of solution)   x 100%

Alternatively,

%w/w = mass of solute (g) in 100 g of solution.

Concentration in grams per litre

Concentration can also be expressed as grams of solute dissolved in one litre of solution.

Example: Preparation of 300 mL of a sucrose solution of concentration 5 g/L.

As only 300 mL of solution is required, only a fraction of the 5 g will be needed. To find the quantity of sucrose required, the concentration is multiplied by the fraction of litres required:

m = 5  g/L  x 0.3 L

= 1.5 g

This amount of sucrose is weighed out and dissolved in enough water to make up the volume to 300 mL.

## Answer by nehal.trivediasta on question Solution making

Hi Patricia,

In simple terms, for the calculation of preparing chemical solution from a solid chemical,

The following information/data are required

1. Molecular weight (in grams) of the chemical solid you are going to use, which should be be available on the label of the container

2. concentration of solution to be prepared

3. Total volume of the solution to be prepared

4. Assay % of chemical solid mentioned in the label (for simplicity I assume it is 100% and will not it mention in the calculation)

Assuming you wish to prepare 0.1M 500 ml copper sulfate

1. concentration - 0.1 Molar copper sulfate as C1

2. volume - 500 ml as V1

3. molecular weight - of copper sulfate pentahydrate solid 249.68 gms as W1

4. weight of the copper sulfate solid required as W2

The following formula will help in the actual calculation -

the required weight W2 of the chemical to be taken for solution preparation is equal to

product of (W1, V1 and C1) divided by 1000

i.e. Formula

W2   =     (W1  X   V1    X    C1 )    /     1000

=  (249.68 X 500  X 0.1)/ 1000

=  12.4850 gms

that means weigh out 12.4850 gms of copper sulfate pentahydrate and dissolve it in distilled water and then make it up to 500ml to make 0.1 molar of copper sulfate solution.

I hope it assists.

Regards

Nehal Trivedi

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