Question

Amount of oxalic acid (COOH)2.2H2O, in grams required to prepare 200mL of 0.5 M oxalic acid solution is

Answer 1

Answer:

If 6.3 g (COOH) 2.2H2O is dissolved in a 500 cm³ solution, then what are its molarity and normality respectively?

This is so simple if you understand the meaning of NORMALITY &MOLARITY . Both terms represents concentration. So , now let's discuss one by one:-

MOLARITY(M) = (number of moles of solute )/( volume of solution ). ————- A

Number of moles can be determined by various ways,but here as we can see that weight of solute is given . So , no. of moles = given weight of solute/molecular weight of solute .

Molecular weight = sum of atomic weight of each atom present in molecule. Ex:-Molecular weight of CO2 = 12+16*2 = 44. NH3 :- 14+ 1*3=17. H2O = 1*2+16 = 18 . COOH.2H2O = 12+16+16+1+2*18= 81 ,( COOH)2.2H2O = 126

NORMALITY(N) = (number of gram equivalent weight of solute)/ (volume of solution).

No. Of gram equivalent weight can be calculated as (Weight of solute)/ equivalent weight of solute .

Equivalent weight =molecular weight of solute / valency.

Ex:- equivalent weight of. CH3COOH= 48/1 = 48.

& for (COOH)2.2H20 = 126/2 = 63 .

So we can write normality in another way as N= (weight of solute * valency) / molecular weight . ————B

We can also derive relation between N and M by comparing equation A and B . i.e N = valency * M.

NOW , Coming to the question ;

Weight of solute is given i.e. 6.3 g

Volume of solution is given i.e. 500 cm3 = 0.5 Litre

So, M = (6.3/126)/0.5 = 0.1

N = Valency * M = 2* 0.1 = 0.2