Question

A sulfuric acid solution containing 571.6 g of H2SO4 per liter of solution has a density of 1.329 g/cm3. Calculate (a) the mass percentage, (b) the mole fraction, (c) the molality, (d) the molarity of H2SO4 in this solution.

Answer 1

Answer:

a) 43%

b) 0.122

c) 7.70 molal

d) 5.83 M

Explanation:

Step 1: Data given

Mass of H2SO4 = 571.6 grams

Density of H2SO4 = 1.329 g/cm³ this means 1.329 grams per 1 mL or 1329 grams per 1L

a) Calculate mass percentage

Mass % = (571.6 grams / 1329 grams)* 100% = 43%

b) Calculate mole fraction

Number of moles H2SO4 = Mass H2SO4 / Molar mass H2SO4

Moles H2SO4 = 571.6 grams / 98.08 g/mol

Moles H2SO4 = 5.83 moles

Moles H2O = (1329 -571.6)/18.02 = 42.03 moles

Total moles = 5.83 + 42.03 = 47.86 moles

Mole fraction H2SO4 = Moles of solute (H2SO4)/ Total moles

Mole fraction H2SO4 = 5.83 moles / 47.86 moles

Mol fraction h2SO4 = 0.122

c) Calculate the molality

Mass of solvent = 1329 grams - 571.6 grams = 757.4 grams = 0.7574 kg

Molality of H2SO4 = number of moles H2SO4 / mass of solvent

Molality H2SO4 = 5.83 moles / 0.7574 kg

Molality H2SO4 = 7.70 molal

d) Calculate Molarity

Molarity H2SO4 = Number of moles H2SO4 / volume

Molarity H2SO4 = 5.83 moles / 1L = 5.83 M