Question

A solution of crystalline oxalic acid is prepared by dissolving 0.63g of
oxalic acid in 250 cc of solution.
Calculate normality and molarity of solution. ​

Answer 1

Answer:

Molar Mass of the Oxalic Acid = 1236 g/mole. = 5 × 10⁻³ moles. Hence, the molarity of the solution is 0.02 M. = 0.04 N.

Explanation:

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Answer 2

Answer:

Given conditions =

ll

Volume of the Solution = 250 cm? = 250 mL.

250 x 10-3 L. =

Mass of the Oxalic Acids = 0.63 grams. Molar Mass of the Oxalic Acid = 1236 g/mole.

For molarity,

No. of moles = Mass/Molar Mass

.: No. of moles = 0.63/126 = 5 x 10-3 moles.

Using the Formula,

Molarity (M) = No. of moles of the Solute/ Volume of the Solution in liter

.: M = (5 x 10-/(250 x 10-)

M = 0.02 M

Hence, the molarity of the solution is 0.02 M.

For Normality,

Basicity of the Oxalic Acid = 2 Normality = Basicity x Molarity

= 2 x 0.02 = 0.04 N.

For Molality,

Molarity calculated = 0.02 M. This means that 0.02 moles of the acid is present in the 1 liter of the solution.

Now, Assuming that the Total number of moles in 1 Liter of the solution is 1.

No. of moles of water = 1 -0.002 moles. = 0.98 moles

Therefore, mass of the water 0.98 x 18 = 17.64 g.

Now,

Molality(m) = No. of moles of solute/Mass of = 0.28 m. =

the solvent or water = (0.005/17.64) x 1000

Hence, molality of the solution is 0.28 m

Explanation:

:-)