Question

Volumetric analysis and their calculations

Answer 1

(1) Strength of solution = Amount of substance in g litre-1

(2) Strength of solution = Amount of substance in g moles litre-1  

(3) Strength of solution = Normality × Eq. wt. of the solute

=  molarity × Mol. wt. of solute

(6) Number of millimoles = Wt. in gm × 1000/mol.wt. 
= Molarity × Volume in Ml. 
(7) Number of equivalents 
=  Wt. in (gm/Eq. wt.) = x × No. of mole × Normality × Volume in litre 
(8) Number of milliequivalents (meq.) 
= (Wt. in gm × 1000/Eq.wt.) = normality × Volume in ml. 
(9) Normality = x× No. of millimoles 
= x× Molarity = (Strength in gm litre-1/Eq.wt.) 
where x = (Mol. Wt./Eq.wt.), x = valency or change in oxi. Number. 
(10) Normality formula, N1V1 = N2V2

Answer 2

1) Strength of solution = Amount of substance in g litre-1

(2) Strength of solution = Amount of substance in g moles litre-1  

(3) Strength of solution = Normality × Eq. wt. of the solute

=  molarity × Mol. wt. of solute

(6) Number of millimoles = Wt. in gm × 1000/mol.wt. 
= Molarity × Volume in Ml. 
(7) Number of equivalents 
=  Wt. in (gm/Eq. wt.) = x × No. of mole × Normality × Volume in litre 
(8) Number of milliequivalents (meq.) 
= (Wt. in gm × 1000/Eq.wt.) = normality × Volume in ml. 
(9) Normality = x× No. of millimoles 
= x× Molarity = (Strength in gm litre-1/Eq.wt.) 
where x = (Mol. Wt./Eq.wt.), x = valency or change in oxi. Number. 
(10) Normality formula, N1V1 = N2V2

(16) Mol. Wt. = V.D × 2 (For gases only)