- Importance Of chemistry
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- Physical States of matter
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- Chemical Classification Of matter
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- The International System of Units (SI units)
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- Significant Figures
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- Laws of Chemical Combination
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- Atomic Mass
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- Molecular Mass
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- Mole Concept (Avogadro Constant) And Molar mass
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- Gram Atomic Mass
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- Gram Molecular Mass
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- Percentage composition
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- StioChiometry
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- Mole Fraction
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- Molarity Defination
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- Molality Defination
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- Normality Formula

where n = number of H

Its unit is N or eq /L .It depends on the chemical reaction being studied and depends on temperature too.

This is used in acid-base reactions and redox reactions

Molarity |
Normality |

number of moles of solute per liter of solution. |
number of gram equivalent of solute per liter of solution. |

Molarity is a measurement of the moles in the total volume of the solution, |
Normality is a measurement of the gram equivalent in relationship to the total volume of the solution. |

Unit is Moles L^{-1} |
Unit is eq L^{-1} |

$Normality = n \times Molarity$

where n = number of H

Calculate the equivalent weight of following:-

(i)H

(ii)NaCl

(iii) NaOH

(iv) Na

i) Molar Mass of H

n = 2 as it has 2 H+ ion

$ Equivalent \; Weight= \frac {Molar \; Mass}{n} = \frac {98}{2} = 49$

ii) Molar Mass of NaCl =23+35.5= 58.5

n = 1

$ Equivalent \; Weight= \frac {Molar \; Mass}{n} = \frac {58.5}{1} = 58.5$

iii)

Molar Mass of NaOH =40

n = 1 as it has 1 OH- ion

$ Equivalent \; Weight= \frac {Molar \; Mass}{n} = \frac {40}{1} = 40$

iv)

Molar Mass of Na

The salt Na

n = 2

$ Equivalent \; Weight= \frac {Molar \; Mass}{n} = \frac {106}{2} = 53$

Calculate the normality of NaOH solution Formed by dissolving 0.2 gm Na OH to make 250 ml solution

Normality(N) = (no.Gram Equivalent of solute)/(Volume of Solution in litre)

No. of Gram Eq. of Solute = weight/Equivalent weight

Now, Equivalent weight=$ \frac {Molar \; Mass}{n} = \frac {23+16+1}{1}=40$

So, N = (No.gram eq.mass)/(Vol (l)

=Weight/(Equivalent weight) × 1000/(V ml)

=(2/40) X (1000/250)

= .2 N

Calculate the number of moles & molarity of N/2 500 ml solution of H

N = n ×M

1/2=2 ×M

1/4=M

Now Molarity of solution

= No of Moles/Volume in litre /> 1/4=n/.5

or n = 1/8

N

where N

N

V

V

Suppose we have mix three solution of same solute of Normality N

Normality of the resulting solution is given by

$N_R= \frac {N_1V_1 +N_2V_2 + N_3V_3}{V_1 + V_2 + V_3}$

Suppose we have mix three solution of different solute(acid) of Molarity M

and no of H

$N_R= \frac {n_1M_1V_1 +n_2M_2V_2 + n_3M_3V_3}{V_1 + V_2 + V_3}$

Find out the volumes of two HCl solutions P (0.5 N) and Q (0.1 N) to be mixed for preparing 2 L of 0.2 N HCl solution

Let y L of P and (2 -y) L of Q are mixed.

N

$0.5 \times y + 0.1 \times (2 -y) = 0.2 \times 2$

$(0.5 -0.1) y = 0.4 -0.2$

$y = 0.5 L$

Therefore,

0.5 L of A and 1.5 L of B should be mixed.

N

Where

N

V

N

V

Molar mass of ionic compound is known as formula Mass i. e. (FM)

So it is the number of Formula mass in gram of solute in 1 litre solution unit = ‘F’ or ‘Moles L

Class 11 Maths Class 11 Physics Class 11 Chemistry