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Find the molality of 15 M H_2SO4 (98.1 g/mol) in a solution with a density of 1.84 g/ml. Assume 1 L of solution.

a. 4.5 mol/kg
b. 8.2 mol/kg
c. 12.6 mol/kg
d. 15.0 mol/kg

1 Answer

3 votes

Final answer:

The calculated molality of the sulfuric acid solution is 40.71 mol/kg. However, this answer does not match any of the given options, suggesting there might be an error in the provided question values or options.

Step-by-step explanation:

To calculate the molality of a 15 M H2SO4 solution with a density of 1.84 g/ml, first we need to find the mass of the solution and the number of moles of the solute (H2SO4).

Assuming 1 L of solution, we have:

  • Mass of the solution = density × volume = 1.84 g/ml × 1000 ml = 1840 g
  • Moles of H2SO4 = molarity × volume = 15 mol/L × 1 L = 15 moles

The mass of the water (solvent) can be found by subtracting the mass of the solute from the total mass of the solution.

  • Mass of H2SO4 = moles × molar mass = 15 moles × 98.1 g/mol = 1471.5 g
  • Mass of water = mass of solution - mass of H2SO4 = 1840 g - 1471.5 g = 368.5 g
  • Mass of water in kilograms = 368.5 g ÷ 1000 = 0.3685 kg

Now we can calculate the molality:

  • Molality = moles of solute / mass of solvent in kilograms = 15 moles / 0.3685 kg = 40.71 mol/kg

However, none of the options given match the calculated molality of 40.71 mol/kg. Please double-check the values provided in the question for accuracy. If you have copied the question correctly, it seems there might be a mistake in the options provided.

User Udi Dahan
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