Tugas Rutin Kimia Part-1

1.  A solution of phosphoric acid was made by dissolving 10.0 g H₃PO₄ in 100.0 mL water. The resulting volume was 104 mL. Calculater the density, mole fraction, molarity, and molality of the solution? Assume water has a density of 1.00 g/cm³ 

Answer:

(1) Density (ρ)

Remember : density of a solution = mass / volume

Mass of H₃PO₄ = 10 gram

Mass of H₂O = ρ x V = 1 g/mL x 100 mL = 100 g

Mass of solution = 100 g + 10 g = 110 g

Volume solution = 104 mL

So, ρ = m/V = 110 g/104 mL = 1,06 g/mL

(2) Mole Fraction (X)

Mol of H₃PO₄ = mass/Mr = 10 g/97,994 g/mol = 0,102 mol

Mol of H₂O = mass/Mr = 100 g/ 18,016 g/mol = 5,551 mol

X(H₃PO₄) = mol of H₃PO₄ / (mol H₃PO₄ + mol H₂O) = 0,102 / (5,551 + 0,102) = 0,018

X(H₂O) = mol H₂O / (mol H₃PO₄ + mol H₂O)= 5,551 / (5,551 + 0,102) = 0,982

(3) Molarity (C)

C = mol/Volume

C(H₃PO₄) = n/V = 0,102 mol/ 0,104 dm3 = 0,98 mol/dm³

(4) Molality (b)

b = mol / mass

b = mol H₃PO₄/ mass H₂O = 0,102 mol/ 0,1 kg = 1,02 mol/kg 

2.  An aqueous antifreeze solution is 40.0% ethylene glycol (C₂H₆O₂) by mass. The density of the solution is 1.05 g/cm³. Calculate the molality, molarity, and mole fraction of the ethylen glycol. 

Answer :

The information we have:

w = 0,40

ρ =1,05 g/cm³

Mass percentage (w) = mass solute / mass solution

And density of a solution = mass / volume

If we substitute the mass, we got :

Mass = w x ρ x V

So, mass of C₂H₆O₂ = w x ρ x V

Mol = (w x ρ x V) / Mr = (0,4 x 1,05 g/cm³ x 1000 cm³)/62,07 g/mol = 6,7666 mol

(1) Molarity (C) = mol/Volume = 6,7666 mol/ 1 dm³ = 6,7666 mol/dm³

Mass of solution = mass of C₂H₆O₂ + mass of H₂O

Mass of solution = (mol of C₂H₆O₂ x Mr C₂H₆O₂) + mass of H₂O

Mass H₂O = mass solution – (mol C₂H₆O₂ x Mr C₂H₆O₂) = ρ x V - (mol C₂H₆O₂ x Mr C₂H₆O₂)

Mass of H₂O = 1.05 g/cm³ x 1000 cm³ – (6.7666 mol x 62.07 g/mol)

Mass of H₂O = 630 g

(2) So, molality (b) = mol/mass = 6,7666 mol/0,63 kg = 10,74 kg

(3) Mole Fraction (X) = mol of C₂H₆O₂ / (mol C₂H₆O₂ + mol of C₂H₆O₂) = 6,7666 / (7,666 + (630/18,01))

So, X = 6,7666 mol/41,65 mol = 0,1625

3.  The lattice energy of Nal is -686 kJ/mol, and the enthalpy of hydration is -694 kJ/mol. Calculate the enthalpy of solution per mole of solid Nal. Describe the process to which this enthalpy change applies. 

Answer :

Solvation of salts can be describe as 2 steps process :

1. breaking of crystal lattice

2. hydration of realeased salt ions

Crystal lattice enthalpy, ∆HLE is defined ,as energy released when forming the crystal lattice. To break the lattice, we need to add energy to overcome this barrier, equal -∆HLE, which is in the case of NaI equal to 686 kJ/mol

Hydration enthalpy, ∆HHyd is energy required to surround the released ions with the solvent molecules, in the case of NaI salvation in water, it is equal to -684 kJ/mol

According to the Hess`s law solvation enthalpy is equal to :

∆HSOLVN = - ∆HLE+ ∆HHyd

∆HSOLVN = 686 kJ/mol – 596 kJ/mol = -8 kJ/mol

4.  Although Al(OH)₃ is insoluble in water, NaOH is very soluble. Explain in terms of lattice energies.

Answer:

Lattice energy is the energy required to separate 1 mol of an ionic solid into gaseous ions

5.  Which solvent, water or carbon tetrachloride, would you choose to dissolve each of the following?

A. KrF₂ = Nonolar = Carbon tetrachloride

B. MgF₂ = Polar = Water

C. SF₂ = Polar = Water

D. CH₂O = Polar = Water

E. SO₂ = Polar = Water

F.  CH₂=CH₂ = Nonpolar = Carbon tetrachloride

Source : slader.com

For number 6-10 read HERE

Thanks. 5NChemistry

Location: Indonesia

Berbagi :