1, 2, 12, solutions  not provided

3.  P0(pure solvent) = 295 torr    
    P(solution) = (0.303 atm/760 torr/atm) = 230.3 torr
    mol fraction solvent = X(solvent) = P/P0
                         = (230.3/295) = 0.7806
             ? g solvent = 0.7806 mols[(59 g/1 mol)] = 42.15 g
    mol fraction solute = (1 - X(solvent),   b/c a two-component mixture 
                        = (1 - 0.7806) = 0.2194
    molality = (moles solute)/(kg solvent) = 0.2194/0.04215 = 5.20 molal

4.  moles(after dilution) = mols(before dilution)
        M*(vol.in L)after = M*(vol.in L)before
                 M(1.0 L) = (0.435)(0.208)
                 M(after) = 0.0905 moles solute per liter
    ? moles Na1+ cations = 0.0905 moles Na2SO4[(2 mol Na1+/1 mol Na2SO4)]
                       = 0.181 moles sodium cations

5.  BCC unit cell:
    i.  contains 2 net barium atoms per unit cell
   ii.  has three atoms in contact along body diagonal (i.e., 4 radii)
        BD = 4r     and     BD = sq.rt.(3)*EDGE 
  iii.  EDGE = cube root(0.1261 nm3) = 0.5015 nm
   iv.  BD = sq.rt.(3)*(0.5015) = 0.8686
         r = BD/4 = 0.217 nm

6.  gather preliminary information...
    i.  ONE LITER of solution has mass = 1,085 g  (via density)
   ii.  ONE LITER of solution has (0.10)(1,085) = 108.5 g solute (via %)
  iii.  ? mol solute = 108.5 g[(1 mol/95 g)] = 1.142 mol MgCl2
   iv.  ? g solvent = (1,085 - 108.5) = 976.5 g solvent
    v.  ? mol solvent = 976.5 g solvent[(1 mol/18 g)] = 54.25 mol H2O
    a.  M = mol solute/Liter solution = 1.142/1.0 = 1.142 mol/L
    b.  X solvent = moles solvent/(moles solute + moles solvent)     
                  = (54.25)/(1.142+54.25) = 0.9794
    c.  b/c this is a two-component mixture,
        X solute = (1.0 - X solvent)= (1.0 - 0.9794) = 0.0206
    d.  m = moles solute/kg solvent = (1.142)/(0.9765) = 1.169
    e.  ? [Cl1-] = 1.142 mol MgCl2/L[(2 mol Cl1-/1 mol MgCl2)]
                 = 2.284 mol/L

7.  a.  ln(P2/P1) = -(delta)H/R[1/T2 - 1/T1]   
        ln(295/142) = -(delta)H/(8.31 J/K-mol)[1/(273+50) - 1/(273+30)]
       -(delta)H = {(0.7311)(8.31)(323)(303)}/(-20) 
        (delta)H =  29.73 kJ/mol
    b.  "normal" boiling point is b.p. when vapor pressure = 1 atm.
        ln(760/295) = -(29,730 J/mol)/(8.31 J/K-mol)[1/T2 - 1/(273+50)]
        -(7.201E-4) = 1/T2 - 1/323
        1/T2 = +3.096E-3 - 7.20E-4 = 2.376E-3 
        T2 = 1/(2.376E-3) = 421 K          (or 148 C)

8.  In a FCC unit cell there are:
    i.  4 net octahedral sites, so 4 net barium cations
   ii.  8 net tetrahedral sites, so 8 net fluoride anions
  iii.  vol.of ONE unit cell = (EDGE)3 = 236.5 Ang.3
        vol.of ONE MOLE unit cells = (236.5)(6.02E+23) = 1.42E+26 Ang.3
        ? cm3 = 1.43E+26 Ang.3[(10-8cm)3/(1 Ang.)3] = 142.4 cm3
   iv.  mass in ONE MOLE unit cells = 4(137.3)+8(19) = 701.2 g
    v.  density = (701.2 g)/(124.4 cm3) = 4.93 g/cm3

9.  gather preliminary information...
    i.  for chloroform
        (a)  ? g = 50 mL[(1.50 g/1.0 mL)] = 75.0 g
        (b)  ? mol = 75.0 g[(1 mol/119.5 g)] = 0.6276 mol
   ii.  for methylene chloride
        (a)  ? g = 75 mL[(1.34 g/1.0 mL)] = 100.5 g
        (b)  ? mol = 100.5 g[(1 mol/85 g)] = 1.182 mol
    a.  mass % = mass chloroform/(total mass solution) 
               = 75/(75+100.5) = 29.94
    b.  mol fraction solute = moles solute/(total moles)
                            = 0.6276/(0.6276+1.182) = 0.347
        b/c this is a two-component mixture,
        mol fraction solvent = (1.0 - X solute) = 1.0 - 0.347 = 0.643
    c.  molarity = moles solute/liter solution
                 = 0.6276/(125/1000) = 5.02 mol/L

10. equilibrium vapor pressure is 108 torr at 50 deg.C  Liquid will be
    present when amt of substance has a calculated v.p.> 108 torr
    a.  calculate vapor pressure, assuming ALL material in gas state
        P(calc'd)(3.0 L) = (1.050g/60 g/mol)(0.082 L-amt/K-mol)(323 K)
        P(calc'd) = 0.3829 atm  (i.e., 291 torr)
        b/c P(calc'd) > P(equilib.), i.e., 291 > 108, liquid is present
    b.  using equilib.P, calculate mass of gas in 3.0 L at 50 deg.C
        (108/760 atm)(3.0 L) = (mass/60 g/mol)(0.082 L-amt/K-mol)(323 K)
                      mass in gas state = 0.9658 g (max.amt gas possible)
    c.  ? g present as liquid = 1.050 g - 0.9658 g = 0.084 g liquid

11. Gather preliminary information...
    i.  mass of ONE LITER solution = 1180 g (from density information)
   ii.  mass solute in ONE LITER = (0.20)(1180) = 236 g (from per cent)
        ? mol solute = 236 g[(1 mol/164 g)] = 1.44 mol solute
        ? g solvent = 1180 - 236 = 944 g solvent  
  iii.  molality solute = moles solute/kg solvent = 1.44/0.944 = 1.52
            ONE mole of Ca(NO3)2, when dissolved, will furnish THREE moles
            of ions b/c it is completely dissociated in solution. So,
        molality ions = 3(molality solute)= 4.56
    a.  f.p. lowering = 1.86(4.56) = 8.5 deg.    
        f.p. temperature = -8.5 deg.C
    b.  b.p. elevation = 0.52(4.56) = 2.38 deg
        b.p. temperature = 102.4 deg.C

13. gather preliminary information...
    i.  ? g benzene = 45.8 mL[(0.879 g/1 mL)] = 40.26 g
   ii.  m = (1.583 g solute/MW solute)/(0.04026 kg benzene)
  iii.  f.p. lowering = 5.53 - 2/866 = 2.664 
        (delta)T = Kf(molality)
           2.664 = (4.90)(1.583/0.0406)/MW
              MW = 72.3 g/mol