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Science Clinic => Chemistry => Topic started by: aabwk4 on Mar 21, 2021

Title: Calculate the mole fraction of H2SO4 in 9.61 M H2SO4. The density of the solution is 1.520 g/mL.
Post by: aabwk4 on Mar 21, 2021

Question 1

Calculate both the boiling point and the freezing point if 46.0 g of glycerol, C₃H₅(OH)₃, is dissolved in 500.0 g of H₂O.

Question 2

Calculate the mole fraction of H₂SO₄ in 9.61 M H₂SO₄. The density of the solution is 1.520 g/mL.

Title: Calculate the mole fraction of H2SO4 in 9.61 M H2SO4. The density of the solution is 1.520 g/mL.
Post by: ndhahbi on Mar 21, 2021

Answer 1

​freezing point = -1.86°C     boiling point = 100.51°C 

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46.0 g/92.09 g/mol = 0.500 mol glycerol

molality of solution = 0.500 mol/0.5000 kg = 1m

boiling point = 100.00°C + (0.51°C/m* 1m) = 100.51°C

freezing point = 0.00°C - (1.86°C/m * 1m) = -1.86°C  

Answer 2

0.231

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Assume 1 L (1000 mL) of solution.

moles H₂SO₄in 1 L = 9.61 mol H₂SO₄

mass of H₂SO₄in solution = 9.61 mol * (98.09g/mol) = 943 g H₂SO₄

mass of 1 L solution = 1000 mL * 1.520 g/mL = 1520. g solution

mass of H₂O in solution = 1520. g - 943 g = 577 g H₂O

moles H₂O in solution = 577 g * (1 mol/18.02 g) = 32.0 mol H₂O

mole fraction H₂SO₄ = 9.61/(9.61 + 32.0) = 0.231

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Title: Re: Calculate the mole fraction of H2SO4 in 9.61 M H2SO4. The density of the solution is 1.520 g/mL.
Post by: mkruger4 on Jul 2, 2022

Thank You