Question

3. 18 gram of oxalic acid is dissolve in 500 ml solution the density of the solution is 0. 95 gram per litre calculate normality molarity molality

Answer 1

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molarity = w×1000/m× v(ml)

= 18×1000/ 90×500

= 2/5 moles/litre

molatilty = 1000M/1000d- M(molar w.)


so, m = 400/950-36

400/914 = 0.437mol/kg
normality= n×Molarity
(for acid n = no. of replacable H+)

= N = 2 ×2/5.
normality = 4/5 answer.

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Answer 2

Answer:

Molarity = 0.4M

Molality = 0.437m

Normality = 0.8N

Explanation:

(i) Molarity:-

Given, the amount of oxalic acid = 18g

The molecular mass of the oxalic acid = 90g/mol

Number of moles of oxalic acid (COOH)₂ $=\frac{18}{90}= 0.2 \;moles$

The volume of the solution = 500ml = 0.5L

The molarity of solution = Moles of oxalic acid/Volume of solution (in L)

Molarity of oxalic acid solution $=\frac{0.2mol}{0.5L} = 0.4\;molL^{-1}$

(ii) Molality:-

Given, the density of solution = 0.95 g/ml

Mass of solution = Density × Volume

Mass of solution  $=0.95\times 500 = 475g$

The mass of the solvent $= 475-18 = 457g = 0.457\;Kg$

The molality of solution = Moles of oxalic acid/Mass of solvent (Kg)

Molality of oxalic acid solution  $=\frac{0.2}{0.457} = 0.437m$

(iii) Normality:-

The acidity of oxalic acid = n- factor = 2

The equivalent weight of oxalic acid = 90/2 = 45 g/eq

The number of gram equivalents = $=\frac{18}{45}=0.4eq$

The normality = No. of gram eq./Volume of solution (Kg)

Normality of solution of oxalic acid $=\frac{0.4}{0.5}=0.8N$

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