Practice problems (Chapters 11 & 12)

1- Arrange the following aqueous solutions in order of decreasing freezing point (all 3 solutions have the same molality m= 0.10 mol/Kg, k f = 1.86 0 C/m) a) NaF, b) C 2 H 5 OH, c) Ni(MnO 4 ) 2 , a)  t f = 0 – t f = 1.86 x 0.10 x i NaF  Na + + F - ( I = 2) t f = - 2 x (1.86 x 0.10) b)  t f = 0 – t f = 1.86 x 0.10 x i C 2 H 5 OH (l)  C 3 H 6 O 3 (aq) ( I = 1) t f = - 1 x (1.86 x 0.10) c)  t f = 0 – t f = 1.86 x 0.10 x i Ni(MnO 4 ) 2  Ni 2+ + 2 MnO 4 - ( I = 3) t f = - 3 x (1.86 x 0.10) Freezing point of: C 3 H 6 O 3 > NaF > Ni(MnO 4 ) 2 2- Arrange the following aqueous solutions in order of increasing boiling point (all 3 solutions have the same molality m= 0.10 mol/Kg, k b = 0.52 0 C/m) a) AlCl 3 , b) BaF 2 , c) CaSO 4 a)  t b = t b - 100 = 0.52 x 0.10 x i AlCl 3  Al 3+ + 3 Cl - ( I = 4) t b = 4 x (1.86 x 0.10) +100 b)  t b = t b - 100 = 0.52 x 0.10 x i BaF 2  Ba 2+ + 2F - ( I = 3) t b = 3 x (1.86 x 0.10) + 100 c)  t b = t b - 100 = 0.52 x 0.10 x i CaSO 4  Ca 2+ + SO 4 -2 ( I = 2) t b = 2 x (1.86 x 0.10) + 100 Boiling point of CaSO 4 < BaF 2 < AlCl 3 3- A solution is prepared by dissolving 24.6 g of C 10 H 15 OH(molar mass= 152 g/mol) in 98.5 g of C 6 H 6 ( molar mass= 78 g/mol) at 25 0 C. Calculate the vapor pressure of this solution ( The vapor pressure of C 6 H 6 is 100.0 mmHg at 25 0 C ) P solution = X solvent x P solvent. P solution = X solvent x 100 X solvent = moles solvent / total moles. Moles of benzene = 98.5 g / 78 g/mol = 1.26 mol of benzene (Slovent) Moles of camphor = 24.6 g / 152 g/mol = 0.16 moles X solvent = 1.26 /( 1.26 + 0.16) X benzene = 0.887 P solution = 0.887 x 100 P solution = 88.7 mmHg 4- Calculate the molarity of an aqueous sugar (C 12 H 22 O 11 ) solution with a molality of 1.22 m and a density of 1.12 g/mL . (molar mass of sugar = 342 g/mol, density of water 1.0 g/mL) Molarity = moles sugar/V(L) solution. in 1Kg of water, there are 1.22 moles sugar mass of sugar = 1.22 mol x 342g/mol= 417.24 g mass of solution = 1417.24 g Volume of solution = 1,417.24g / 1.12g/mL = 1265.39 mL = 1.265 L . molarity = 1.22 / 1.265 molarity = 0.96 M 5- Calculate the number of moles of LiCl in a 1.5 L LiCl(aq) solution with an osmotic pressure of 2.45 atm. at 25 0 C ( Molarity = moles solute / V(L) of solution)

 = iMRT M =  / i RT LiCl(s)  Li + + Cl - (i = 2) . M = 2.45/ 2 x 0.0821 x 298. Molarity = 0.05 atm. Moles LiCl = 0.05 x 1.5 0.075 moles LiCl 6- 5.00 mg of an ionic compound with a Vant-Hoff factor of 2 ( I = 2) is dissolved in water to make a 50.00 mL aqueous solution. The solution shows an osmotic pressure of 9.2 mmHg at 42 0 C, what is the molar mass of the ionic solid?  = iMRT M =  / i RT M = (9.2/760)/ 2 x 0.0821 x 315 Molarity = 0.000234 mol/L moles = M x V(L) , moles of solid = 0.000234 x 0.050, moles of ionic solid = 0.000117 MOLAR MASS = mass(g)/moles MOLAR MASS = 0.005/0.000117 MM = 42.73 g/mol 7- Determine the intermolecular forces in: a) H 2 CO 3 b) CH 4 c) H 2 S H 2 CO 3 CH 4 H 2 S London dispersion: X X X Dipole-dipole: X X H-bonding: X X 8- Which of the following species are capable of hydrogen-bonding with H 2 O? a) CO b) HF c) HI CO HF HI : C= O: H-F: H-I: YES YES NO 9- How much heat is required to convert 50.0 g of water at 25 0 C to steam at 150 0 C ? ( t boil = 100 0 C, s w (H2O(l)) = 4.184 j/g . o C,  H vap = 40.67 kj/mol, s w (H2O(g)) = 2.03 j/g . o C) Heat H 2 O(l) from 25 to 100 0 C q 1 = (50). (4.184) . (100 – 25) q 1 = 15.7 kj H 2 O(l)  H 2 O(g)  H vap = 40.67 kj/mol  H vap = 113 kj Heat H 2 O(g) from 100 to 150 0 C q 2 = (50). (2.03).(150 – 100) q 2 = 5.075 kj Total energy needed to complete this process is 133.77 kj 10- Calculate the amount of heat (in kj) released to completely convert 7.35 g of Iodine I 2 (g) to liquid (condensation). Molecular Iodine I 2 has a molar mass of 253.8 g/mol and  H vaporization = 47.03 kj/mol I 2 (g)  I 2 (l)  H cond = - 47.03 kj/mol Energy = -47.03 x 7.35/253.8 Energy = -1.36 kj