3.75M sulfuric acid solution that has a density of 1.23g/ml. Calculate the mass percent,molality and normality of sulfuric acid?H=1g/mol , O=16g/mol , S=32 g/mol

Md Ariful Islam

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Solution:

sulfuric acid = H2SO4

The molar mass of H2SO4 = 2×Ar(H) + Ar(S) + 4×Ar(O) = 2×1 + 32 + 4×16 = 98 (g/mol).

Suppose that the volume of the solution is 1 L (1000 mL).

Thus,

Mass of 1 L H2SO4 solution = Solution volume × Density = 1000 mL × 1.23 g/mL = 1230 g

Moles of H2SO4 = Molarity of H2SO4 × Solution volume = 3.75 mol/L × 1 L = 3.75 mol

Mass of H2SO4 = Moles of H2SO4 × Molar mass of H2SO4 = 3.75 mol × 98 g/mol = 367.5 g

Mass of solvent = Mass of 1 L H2SO4 solution - Mass of H2SO4 = 1230 g - 367.5 g = 862.5 g = 0.8625 kg

Equivalent mass of H2SO4 = Molar mass of H2SO4 / Basicity of H2SO4 = 98 g mol-1 / 2 = 49 g/mol

Hence,

(1):

Mass %H2SO4 = (Mass of H2SO4 / Mass of solution) × 100%

Mass %H2SO4 = (367.5 g / 1230 g) × 100% = 29.88%

The mass percent of H2SO4 is 29.88%

(2):

Molality of H2SO4 = Moles of H2SO4 / kg of solvent

Molality of H2SO4 = 3.75 mol / 0.8625 kg = 4.3478 mol/kg = 4.35 mol/kg

The molality of H2SO4 solution is 4.35 mol/kg

(3):

Normality of H2SO4 = (Mass of H2SO4) / (Equivalent mass of H2SO4 × Solution volume)

Normality of H2SO4 = (367.5 g) / (49 g/mol × 1 L) = 7.5 N

The normality of H2SO4 solution is 7.5 N

or

Since H2SO4 has 2 replacible H, n-factor = 2.

Normality of H2SO4 = Molarity × n-factor = 3.75 × 2 = 7.5 N

Answer:

The mass percent of H2SO4 is 29.88%

The molality of H2SO4 solution is 4.35 mol/kg

The normality of H2SO4 solution is 7.5 N

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