Calculate the weight in grams of the primary standard solid oxalic acid needed to prepare 250 ml of a 0.200 n solution to be used in this laboratory exercise.

2.25 g.

Explanation:

N = (mass / equivalent mass) of the solute x (1000 / V of the solution)

The solute is oxalic acid.

N is the normality of the solution (N = 0.2 N),

mass of the solute (oxalic acid) (??? needed to be calculated),

equivalent mass of the oxalic acid = molar mass / n (the no. of reproducible H) = (90.03 g/mol) / (2) = 45.015.

V is the volume of the solution (V = 250 ml).

∴ The mass of the oxalic acid in g = (N x equivalent mass x V) / 1000 = (0.2 N) (45.015) (250.0 ml) / 1000 = 2.25 g.

Very important note:

If the solute source is oxalic acid dihydrate:

The molar mass = 126.07 g/mol.

The equivalent mass = 63.035 g.equ./mol.

∴ The mass of the oxalic acid in g = (N x equivalent mass x V) / 1000 = (0.2 N) (63.035) (250.0 ml) / 1000 = 3.15 g.

answer

the ions in solute supperate from eachouthere

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