How to find out the Mass of Salt required to achieve the required Concentration of a Solution

NOTE: E&OE, means errors and omissions are expected and excepted, in other words, if there is any error in the following ...

NOTE: E&OE, means errors and omissions are expected and excepted, in other words, if there is any error in the following article, the same has to be corrected by the user or the beneficiary accordingly.
This article contains works derived from other reputed works of huge publication houses and notes that were made by me in my school life.These articles are just to help people who need these articles desperately and urgently. Any queries, complaints, and corrections will be enthusiastically entertained.

At times, the questions in a Board Practical exams the mass of the solute or the salt may not be given, which directly indicates for you to figure the weight by yourself.

So, the basic formula to work with is based on two concentrations terms, that are namely molarity and normality.

MOLARITY:

Molecular mass of the salt = x

Therefore, for preparing 1000ml of 1M of the given salt, weight of the salt required = m

Hence, for preparing the solution with the required amount(a) and of the concentration mentioned (c), mass of salt required = (mx1000)x250x(c)

Simply, the formula stands, Molarity = Mass of solute (in grams) per litre of solution / Gram molecular mass of the solute

For example-

Here, we have to find the required salt to prepare 250ml of M/20 solution of Mohr’s salt.

Molecular mass of Mohr’s salt, FeSO₄.(NH₄)2SO₄.6H₂O = 392

Hence, for preparing 1000ml of 1M Mohr’s salt solution,

Mohr’s salt required = 392gm

Therefore, for preparing 250ml of M/20 Mohr’s Salt Solution,

Mohr’s Salt required = (392/1000)x250x(1/20) = 4.9gm

NORMALITY:

By studying the equation of the redox reaction that occurs, determine the number of electrons that one loses, that will be denoted by (n).

Therefore, Equivalent mass of the salt (eq. m) = Molecular mass of the salt / Number of electrons lost by one molecule of it.

Strength (s)(gm/l) = Normality x Equivalent mass = (N x eq. m) gm/l (N is the specified normal concentration in the question)

Therefore, for preparing 1 litre of N normal salt solution s gm of it to has to be dissolved.

Therefore, for preparing required amount of solution (a) of N normal salt solution, the requred salt crystals are = (s/1000) x a

Simply, the formula stands, Normality = Mass of solute (in grams) per litre of solution / Gram-equivalent mass of the solute

For example-

Here, we have to find the required salt to prepare  250ml of N/20 solution of Mohr’s salt.

The molecular formula of Mohr’s salt is FeSO₄.(NH₄)2SO₄.6H₂O.

The ionic equation for the ionic equation for the oxidation of Mohr’s salt is, Fe²⁺à Fe³⁺ + e^-

By studying the above equation of the redox reaction that occurs, we get the number of electrons that iron or ferrous loses, which is (n) = 1.

Therefore, Equivalent mass of Mohr’s salt (eq. m) = Molecular mass of Mohr’s salt / Number of electrons lost by one molecule of it. = 392/1 = 392

Strength (s)(gm/l) = Normality x Equivalent mass = (1/20) x 392 gm/l = 19.6 gm/l

Therefore, for preparing 1 litre of N/20 salt solution 19.6 gm of it to has to be dissolved.

Therefore, for preparing 250ml of N/20 normal salt solution, the required salt crystals are = (19.6/1000) x 250 = 4.9g

Some molecular masses and equivalent masses of some substances are given below, which will eventually help you in calculations:-

Substance	Molecular Mass	Ionic Equation	n	Equivalent Mass
Potassium Permanganate (KMnO4)	158	MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O	5	31.6
Mohr's Salt [FeSO4(NH4)2SO4.6H2O]	392	Fe2+ --> Fe3+ + e-	1	392
Ferrous Sulphate (anhydrous) (FeSO4)	152	Fe2+ --> Fe3+ + e-	1	152
Ferrous Sulphate (crystals) (FeSO4.7H2O)	278	Fe2+ --> Fe3+ + e-	1	278
Oxalic acid (anhydrous) (H2C2O4)	90	C2O42- --> 2CO2 + 2e-	2	45
Oxalic acid (crystals) (H2C2O4.2H2O)	126	C2O42- --> 2CO2 + 2e-	2	63

So the basic concentrations by factor that are given are:

y = M molar and N normal

• y/10 = 0.1
• y/20 = 0.05
• y/50 = 0.02
• y/100 = 0.01

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To know more click on the articles below (Important for Boards):-

• To prepare 250ml of M/20 standard Mohr's Salt solution.
• Apparatus Used in Volumetric Analysis
• Some important terms related to concentrations of solutions
• Equivalent masses of oxidizing and reducing agents
• Preparing a standard solution
• Viva Questions related to Volumetric Analysis

Reference:-

• Verma, Dr. N.K.. Comprehensive Practical Chemistry, Class XII.
• NCERT. Chemistry Part I and II, Class XII.

For latest syllabus, both theory and practical papers for class XII, check out the CBSE’s official site http://cbseacademic.in/ for more info.

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Related Questions-

How to calculate molarity

How to calculate normality

How to calculate strength

How to find out the mass of salt required to achieve the required concentration of a solution

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