dilution problem ; chemistry

Question

An aqueous solution of an unknown acid of the formula HX has a mass percentage composition of 2.56% and a mass density of 1.10 g/ml. 10 ml of this solution is taken and water is added until a total volume of 500 ml is reached. The concentration in mol/L of the resulting dilute solution of HX is 0.0045 mol/L. The identity of X is:

Answer 1

0.0045 mol / L x 0.500 L = 0.00225 moles HA

This is the # moles present in 10 ml of original solution. Thus, moles / ml = 0.00225 mol / 10 ml = 0.000225 moles HA / ml original solution

Original soln:

1.10 g / ml x 2.56% HA = 0.02816 g HA / ml

molar mass HA = g / mole = 0.02816 g / 0.000225 moles = 125 g / mol = molar mass

H = 1

X = 124

X = I (atomic mass 127, close enough)

Identity of X is probably iodine, unless you want to consider some organic acids. If so, it could be benzoic acid (molar mass 122),