Answer to Question #253118 in Inorganic Chemistry for chin

Question

A Mixture Contains 1.5 Liters Oxalic Acid H2C2O4· 2H2O (ρ=1.653 g/mL) and 2-

gallon Water. Determine:

ρ Oxalic acid = 1.653 g/mL

ρ water = 1000 g/L

a) %w/w, %w/v, %v/v of oxalic acid in a solution (6 pts)

b) ppm and ppb of oxalic acid (4 pts)

c) Molarity, Normality, and molality of the solution (10 pts)

Answer 1

Molecular weight of oxalic acid (H2​C2​O4.2H2O) is 126 g/mol.

Molarity of 0.2 N oxalic acid (H2​C2O4.2H2O) solution =$\frac{0.2​}{2}=0.1 M$

Amount of solution =50 mL =0.05 L

So, amount of oxalic acid =126×0.1×0.05=0.63 g