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Chapter 13: Problem 47

A sulfuric acid solution containing 571.6 \(\mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) per liter of solution has a density of 1.329 \(\mathrm{g} / \mathrm{cm}^{3} .\) Calculate (a) the mass percentage, (b) the mole fraction, (c) the molality, ( \mathbf{d} ) ~ t h e ~ m o l a r i t y ~ o f ~ \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in this solution.

Short Answer

Expert verified
(a) The mass percentage of H2SO4 in the solution is approximately 43%. (b) The mole fraction of H2SO4 in the solution is 0.122. (c) The molality of H2SO4 in the solution is 7.69 mol/kg. (d) The molarity of H2SO4 in the solution is 5.83 mol/L.

Step by step solution

01

(a) Calculate the mass of 1 liter of solution

Given that the density of the solution is 1.329 g/cm³, we need to find the mass of 1 liter of the solution. We know that 1 liter is equal to 1000 cm³. Now we can use the density formula mass = volume × density to calculate the mass of 1 liter of solution: mass = 1000 cm³ × 1.329 g/cm³ = 1329 g.
02

(a) Calculate mass percentage of H2SO4 in the solution

Now that we know the mass of 1 liter of the solution (1329 g) and the mass of H2SO4 in 1 liter of the solution (571.6 g), we can calculate the mass percentage using the formula: mass percentage = (mass of solute / mass of solution) × 100: mass percentage = (571.6 g / 1329 g) × 100 = 43% (approximately).
03

(b) Calculate the moles of H2SO4 and moles of solvent

To calculate the mole fraction, we need to find the moles of H2SO4 and moles of solvent. First, let's find the moles of H2SO4 using its given mass and molar mass: moles of H2SO4 = mass of H2SO4 / molar mass of H2SO4 = 571.6 g / 98.08 g/mol = 5.83 mol. Next, we need to find the moles of solvent (water). First, let's find the mass of water: mass of water = mass of solution - mass of H2SO4 = 1329 g - 571.6 g = 757.4 g. Now, let's find the moles of water using its molar mass (18.015 g/mol): moles of water = mass of water / molar mass of water = 757.4 g / 18.015 g/mol = 42.06 mol.
04

(b) Calculate the mole fraction of H2SO4

Now that we have the moles of H2SO4 and water, we can calculate the mole fraction of H2SO4 using the formula: mole fraction = moles of solute / (moles of solute + moles of solvent): mole fraction of H2SO4 = 5.83 mol / (5.83 mol + 42.06 mol) = 5.83 mol / 47.89 mol = 0.122.
05

(c) Calculate the molality of H2SO4

To calculate molality, we need to find the moles of the solute (H2SO4) and the mass of the solvent (water) in kilograms. We already have the moles of H2SO4 (5.83 mol) and the mass of water (757.4 g). Let's convert the mass of water to kilograms: mass of water = 757.4 g × (1 kg / 1000 g) = 0.7574 kg. Now, let's find the molality by using the formula: molality = moles of solute / mass of solvent (in kg): molality = 5.83 mol / 0.7574 kg = 7.69 mol/kg.
06

(d) Calculate the molarity of H2SO4

Lastly, we need to find the molarity of H2SO4. To do this, we need to know the moles of H2SO4 and the volume of the solution in liters. We know that there are 5.83 moles of H2SO4 and 1 liter of the solution. Now, we can find the molarity using the formula: molarity = moles of solute / volume of solution (in L): molarity = 5.83 mol / 1 L = 5.83 mol/L. #Results# (a) The mass percentage of H2SO4 in the solution is approximately 43%. (b) The mole fraction of H2SO4 in the solution is 0.122. (c) The molality of H2SO4 in the solution is 7.69 mol/kg. (d) The molarity of H2SO4 in the solution is 5.83 mol/L.

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