Question

1. List the following aqueous solutions in order of increasing boiling point: 0.120 m glucose, 0.050m LiBr,0.050m Zn(NO3)2

2. we are told that the vapor pressure of ethanol (CH3CH2OH) at 20C is 45 torr. The vapor pressure of methanol (CH3OH) at 20C is 92 torr. Calculate thevapor pressure of a solution which contains 25 g of methanol and 75 g of ethanol.

Answer 1

, glucose is a molecular compound, and therefore a nonelectrolyte. a 0.120 m glucose solution simply has 0.120 mol glucose molecules / kg solvent.

LiBr is a salt, and therefore a strong electrolyte. LiBr in aqueous solution ionizes completely:

LiBr(s) -----> Li+(aq) + Br-(aq)
each mol of LiBr produces 2 mol of ions, so the total number of moles of ions in solution is

mom LiBr=1. , mom ions = 1

0.050 mom LiBr/kg * 2 mom ions / mol LiBr = 0.10 mol ions / kg

so 0.050 m LiBr has fewer particles in solution than does 0.120 m glucose (0.1 m vs 0.120 m), and therefore boils at a lower temperature than the glucose solution.

zinc nitrate, Zn(NO3)2, is also a salt which ionizes according to

Zn(NO3)2(s) -----> Zn2+ + 2NO3-(aq)

so each mol of zinc nitrate produces 3 moles of ions. therefore, our 0.05 m zinc nitrate solution has

mom ZnNo3=1
0.050 mol ZnNo3/kg *3 mol ions /mol ZnNo3 = 0.15 mol ions /kg

or 0.15 mol of particles per kg. This is more particles than both the glucose and LiBr solutions.

therefore, the zinc nitrate has the highest boiling point; the glucose is a bit lower, and the LiBr solution has the lowest boiling point.

2)

calculate the mole fraction of methanol and ethanol in the solution:

25 g methanol / 32.0 g/mol = 0.7813 mol methanol
45 g ethanol / 46.1 g/mol = 0.9761 mol ethanol

mole fraction methanol = 0.7813/ (0.7813 + 0.9761) = 0.445
mole fraction ethanol = 1 - 0.445 = 0.555

Vapor pressure methanol = 92 mm Hg X 0.445 = 40.94 mm Hg

Vapor pressure ethanol = 44 mm Hg X 0.555 = 24.42 mm Hg

Vapor pressure of solution = 40.94+ 24.42 = 65.36 mm Hg ..

Thank u..