What is the normality and nature of a mixture obtained by mixing 0.62 g of Na2CO3 •H2O to 100mL of 0.1N H2SO4?

Answer 1

I don't understand the problem.
Na2CO3 + H2SO4 ==> Na2SO4 + H2O + CO2

There is the equation. I have no idea what "nature of the mixture" means but let's see what happens.
How many grams Na2CO3 will the H2SO4 use? That's ml x N x milliequivalent weight Na2CO3 = grams Na2CO3.
100 mL x 0.1N x 0.053 = approx 0.52 but you need a more accurate answer than that.
You had 0.62 g Na2CO3 initially, you have used 0.52 which leaves about 0.1 g Na2CO3 unused. (Remember to recalculate all of these numbers.) The normality of that solution will be N = eq/L
equivalents = grams/eq weight = 0.1/53 = ?
N = eq/L = ?eq/0.1L = ?
So the nature of the solution may be that the H2SO4 is completely used and you have a solution of Na2CO3 in the liquid and the N of the Na2CO3 solution is ?? N.