Question

A solution of oxalic acid is prepared by dissolving 0.63g of acid in 250cm cube of the solution .calculate molarity, molality and normality

Answer 1

Final Answer : Molarity : 0.02M
Normality : 0.04N


Steps:
1) Oxalic Acid: C2H2O4.(2H2O )
Molar Mass : 126g
Given Mass of Oxalic acid /solute = 0.63g
no. of moles of solute = 0.63/126 = 0.5 *10^(-2)

Volume of solution, V = 250cm^3 = 0.25L
=> Molarity, M
= no. of moles of solute / Volume of solution
$= \frac{0.5 \times {10}^{ - 2} }{0.25} = 2 \times {10}^{ - 2}$

Hence,Molarity = 0.02M

2) Since, Basicity / Valency Factor of acid is 2 .(Two ionisable H)
=> Normality =Molarity * Valency Factor
= 0.02M * 2 = 0.04N.

3) For Molality, Data is insufficient.
As We can't get Weight of solvent in solution from required data.

Answer 2

Given conditions ⇒ 

Volume of the Solution = 250 cm³. 
= 250 mL.
= 250 × 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 × 10⁻³ moles. 

Using the Formula,

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

∴ M = (5 × 10⁻³)/(250 × 10⁻³)
   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 the solvent or water
= (0.005/17.64) x 1000
= 0.28 m.

Hence, molality of the solution is 0.28 m



Hope it helps. :-)