equation for calculating normality with example

Let, Mass of the organic compound = W g Volume of the standard acid required for complete neutralization of the evolved ammonia = V mL Normality of the standard solution of acid = N From the law of equivalence (normality equation), 1000 mL of 1 N acid = 1000 mL of 1 N..

The normality equation is commonly used to calculate the normality of solutions after dilution. The above equation is also called as dilution formula because it helps in calculating the volume of the solvent required in diluting a co..

To calculate the volume of a definite solution required to prepare solutions of other normality, the following equation is used: N 1 V 1 =N 2 V 2 , where N 1 =initial normality and N 2 =normality of the new solution, a nd V 1 = initial volume and V ..

Normality of a solution is defined as the number of gram equivalents (gm.e) of a solute dissolved per litre of the given solution. Mathematically it is, Normality is expressed by the symbol N. It can also be expressed as,..

The normal at any point of the spherical mirror is the straight line obtained by joining that point with the centre of the mirror. The normal at point A on the mirror is the line AC obtained by joining A to the centre of curvature of the mirror. Normal at any point on..

Normality is defined as the number of gram equivalents of solute present per litre (dm 3 ) of the solution at any given temperature. It is expressed as N. In terms of weight, Normality of the substance can be expressed as:..

Mass of U-tube with contents before experiment = a grams Mass of U-tube with contents after experiment = b grams Hence, mass of water vapour absorbed = (b - a) g Volume of air = x litres..

Let, The mass of the compound taken be = W g Mass of AgX formed = W 1 g From stoichiometry, Ag = X 1 mol 1 mol 1 molecular mass 1 atomic mass Therefore, ..

Let, The mass of the organic compound taken be = W g Mass of Mg 2 P 2 O 7 obtained = W 1 g From stoichiometry, Mg 2 P 2 O 7 = 2P 1 mol 2 mol 1 x molecular mass 2 x atomic mass 222 g 2 x 31 g = 62 g Therefore, Then, ..

Let the mass of platinic chloride salt be = W g The mass of platinum residue left = x g From the molar stoichiometry, Number of mol of B 2 H 2 PtCl 6 = Number of mol of Pt Or If E is the equivalent mass of the organic base, Then molecular mass of the salt (B 2 H 2 PtCl 6 ) = 2E + 2 + 195 + 213 ..