Aakash ANTHE | Click Here |

NTSE | Click Here |

KVPY | Click Here |

N factor here we mean a conversion factor by which we divide molar mass of substance to get equivalent mass and it depends on nature of substance which varies from one condition to another condition. We can divide n-factor calculation is two category.** **

**In case of non-redox reaction.**

**(a) ****n factor of acid** =Basicity of the acid

**Basicity:** Number of replaceable H^{+} ion.

**Ex.1**

n factor of HCl = 1

n factor of CH_{3}COOH=1

n factor of H_{2}SO_{4}=2

**(b)****n factor of base=** acidity of the base

**Acidity: **Number of replaceable OH^{–} ion.

** **

**Ex.2 **

n factor of NaOH=1

n factor of Ca(OH)_{2}=2

n factor of Al(OH)_{3}=3

n factor of B(OH)_{3}=1 (because it is a mono basic acid)

**(c) ****n factor Salt:** Total number of positive or negative charge.

** **

**Ex.**3

n factor of NaCl=1

n factor of Na_{2}SO_{4}=2

n factor of K_{2}SO_{4}. Al_{2} (SO_{4})_{3}. 24H_{2}O = 8

** **

**Ex.4 **

Find the n factor of H_{3}PO_{4} in the following reaction.

H_{3}PO_{4} + Ca (OH)_{3} → CaHPO_{4} + 2H_{2}O

**Sol.**

Basicity of H_{3}PO_{4} in the above reaction is 2

∴ the n factor of H_{3}PO_{4} is 2

** **

**In case of redox reaction**

**(a) ****From oxidation number**

n factor of oxidizing or reducing agent=change in oxidation number per molecule.

⇒ consider a salt A_{x}B_{y} in which the O.S. of A is +c. It changes to a compound A_{d}E in which the O.S. of A is +f. Here we are assuming that B does not undergo any change in O.S. A^{+c}_{x}B_{y} → A^{+f}_{d}E (Obviously A_{x}B_{y} must have reacted with some other substance to produce the product A_{d}E. That means other substance has the atom E in it.) The ‘n’ factor is = |xc-xf|.

** **

**Ex.5**

Find the n factor of KMnO_{4} in different medium.

**Sol.**

(i) In acidic medium

KMnO_{4}→Mn^{+2
}Change in oxidation number of Mn=+7-2=5

∴ the n factor of KMnO_{4}=5

(ii) Basic medium

KMnO_{4}→K_{2}MnO_{4
}N factor of KMnO_{4} = +7 – 6 = 1

(iii) Neutral medium

KMnO_{4} → MnO_{2
}N factor of MnO_{2} = +7 – 4 = 3

** ***(b)From ion electron method:*

N factor = total number of electrons transferred per mole of the reactant

« Click Here for Previous Topic | Click Here for Next Topic » |

Click Here for Class XI Classes Chemistry All Topics Notes