An aqueous solution of sulfuric acid is 2.88 molal and has a density of 1.7678 g

Question

An aqueous solution of sulfuric acid is 2.88 molal and has a density of 1.7678 g mL-1. Calculate the a) weight percent sulfuric acid; b) mol fraction of sulfuric acid; c) the molarity of sulfuric acid. PS5.8. A solution of glycerol, C3HaOs, in water is prepared by mixing 168.1 g C3H8O3 with 1278.5 g of water. The molarity was found to be 1.301 M. Calculate a) the molality of the glycerol solution b) the density of the glycerol solution PS5.9. A solution of a compound called potassium acid phthlate, is prepared by dissolving 2.34 g in 200 mLs of water. This solution is then reacted with exactly 15.78 mLs of a 0.726 M NaOH solution. Assume KHP reacts in a one-to-one ratio with NaOH, calculate the molar mass of KHP (potassium acid phthlate)

Explanation / Answer

Solution :-

Given aqueous sulfur ic acid molality = 2.88 and density = 1.7678 g/ml

Lets assume we have 1 kg water = 1000 g

Then moles of H2SO4 present = 2.88 mol

Now lets calculate the mass of H2SO4

Mass of H2SO4 = 2.88 mol * 98.079 g per mol = 282.5 g

Total mass of solution = 1000 g water + 282.5 g H2SO4 = 1282.5 g

a). Now lets calculate the weight percent of sulfuric acid

Weight % =( mass of sulfuric acid /total mass)*100%

                 = (282.5 g / 1282.5 g)*100%

                = 22.03 %

b) mole fraction of sulfuric acid

moles of water = 1000 g / 18.0148 g per mol = 55.51 mol

total moles = 2.88 mol H2SO4 + 55.51 mol water = 58.39 mol

mole fraction of H2SO4 = moles of H2SO4 / total moles

                                           = 2.88 mol / 58.39 mol

                                          = 0.0493

c) molarity = moles / liter

lets calculate the volume of solution

volume = mass / density

               = 1282.5 g / 1.7678 g / ml

               = 725 ml

725 ml * 1 L / 1000 ml = 0.725 L

Molarity = 2.88 mol / 0.725 L

                = 3.97 M

PS5.8

Given glycerol = 168.1 g

Water = 1278.5 g

Molarity = 1.301 M

Lets calculate the moles of glycerol

Moles of glycerol = mass / molar mass

                                = 168.1 g / 92.09382 g per mol

                                = 1.8253 mol

Now lets calculate the volume of the solution

Volume = moles / molarity

              = 1.8253 mol / 1.301 mol per L

               = 1.40 L

a) molality =moles / kg solvent

                    = 1.8253 mol / 1.2785 kg

                   = 1.428 m

b) density = mass / volume

1.40 L = 1400 ml

                   = (168.1 g + 1278.5 g)/ 1400 ml

                 = 1.0333 g / ml

PS5.9

Mas of KHP = 2.34 g

Lets calculate the moles of the NaOH

Moles of NaOH = moalrity * volume inliter

                              = 0.726 mol per L * 0.01578 L

                             = 0.011456 mol NaOH

Mole ratio of the NaOH and KHP is 1 : 1

So the moles of KHP reacted = 0.011456 mol

Now lets calculate the molar mass of KHP

Molar mass = mass/ moles

                      = 2.34 g / 0.011456 mol

                  = 204.25 g per mol

So the molar mass of KHP is 204.25 g per mol

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