Question

concentrated sulphuric acid is approximately 18 molar. 5 cc of it is added to make 500 cc of the solution. the approximate normality of solution will be.... Given options - a. 0.18 , b. 0.09 , c. 0.36 , d. 0.27 And I know that the option is "c" but what's the procedure...

Answer 1

by law of equivancy,

N1×V1=N2×V2.

N1=2×18.
V1=5cc.
V2=500cc.
N2=?

36×5=500×N2,
N2=0.36

Answer 2

Find the normality of sulphuric acid with molarity 18 molar ...

Explanation:

Normality

It is defined as : number of gram equivalents of solute dissolved per litre of solution

Its formula is : N=Number of gram equivalents of solute /volume of  solution

Its unit is : gm equivalent/litre

we know ,

Normality = number of gram equivalents /volume of solution

Number of gram equivalents = Mass of solute /Eq.wt. .

Eq wt . =Molar mass / basicity of an acid

we do have relationship between normality and molarity :

Normality = molarity x molar mass/eq.wt.

For acids : Normality = molarity x basicity

So, in above question :

Molarity of solution = 18 molar

volume  of solute = 5 cc =5/1000=.005L

volume of solution = 500 cc=500/1000=.5L

Molar mass of sulphuric acid = 1 x 2 + 32+16 x 4 = 2+32+64=66+32=98g

Normality of solution = 18 x 2 =36

Volume of solute =5 cc

Normality of solution =?

Volume of solution =500

We can equate them as :

N₁V₁=N₂V₂

36 x 5=N₂ x 500

or,N₂=36 x 5 /500=0.36