1) Where would you expect N2 to fit on this graph? a) Its curve will be just bel

Question

1) Where would you expect N2 to fit on this graph?

a) Its curve will be just below that of He

b) Its curve will be just above of that of CH4

c) Its curve will be just above that of O2

d) Its curve will be just below that of CO

2) In Equation Hsoln=H1+H2+H3 which of the energy terms for dissolving an ionic solid would correspond to the lattice energy? Check all that apply.

-delta H1

-delta H2

-delta H3

-delta Hsoln

3) Which aqueous solution will have the lowest freezing point?

a) 0.20 m C12H22O11

b) 0.15 m NaCl

c) 0.10 m HCl

d) 0.050 m CaCl2

e) 0.050 m CH3COOH

4) A mysterious white powder could be powdered sugar (C12H22O11), cocaine (C17H21NO4), codeine (C18H21NO3), norfenefrine (C8H11NO2), or fructose (C6H12O6). When 82 mg of the powder is dissolved in 1.50 mL of ethanol (d = 0.789 g/cm3, normal freezing point 114.6 C, Kf = 1.99 C/m), the freezing point is lowered to 115.5 C. What is the identity of the white powder?

a) powdered sugar

b) codeine

c) cocaine

d) fructose

e) nonfenefrine

5) Proteins frequently form complexes in which 2, 3, 4 or even more individual proteins (“monomers”) interact specifically with each other via hydrogen bonds or electrostatic interactions. The entire assembly of proteins can act as one unit in solution, and this assembly is called the “quaternary structure” of the protein. Suppose you discover a new protein whose monomer molar mass is 25,000 g/mol. You measure an osmotic pressure of 0.0916 atm at 37 C for 7.20 g of the protein in 10.00 mL of an aqueous solution. How many protein monomers form the quaternary protein structure in solution? Treat the protein as a nonelectrolyte.

a) 1

b) 2

c) 3

d) 4

e) 8

6) List the following aqueous solutions in order of decreasing freezing point: 0.040 m glycerin (C3H8O3), 0.020 m KBr, 0.030 m phenol (C6H5OH). Rank solutions from highest freezing point to lowest freezing point. To rank items as equivalent, overlap them.

7) The vapor pressure of benzene, C6H6, is 100.0 torr at 26.1 C. Assuming Raoult’s law is obeyed, how many moles of a nonvolatile solute must be added to 100.0 mL of benzene to decrease its vapor pressure by 10.0% at 26.1 C? The density of benzene is 0.8765 g/cm3.

a) 0.11237

b) 8.765

c) 0.011237

d) 0.1248

e) 0.1282

8) Consider two solutions, one formed by adding 10 gof glucose (C6H12O6) to 1 L of water and the other formed by adding 10 g of sucrose (C12H22O11) to 1 L of water.

Part A)

Calculate the vapor pressure for the first solution at 20 C. (The vapor pressure of pure water at this temperature is 17.5 torr.)

Part B)

Calculate the vapor pressure for the second solution at 20 C. (The vapor pressure of pure water at this temperature is 17.5 torr.)

Explanation / Answer

(3). The depression in freezing point, DeltaTf of a homogenous solution can be calculated from the following formulae

DeltaTf =ix Kf x m

where Kf is the molal depression in freezing point constant of water, and Kf = - 1.86 DegCKgmol-1 . Since all are aqueous solution,the value of Kf remains constant for all solution.

m = molality of the solution

i = Vant-hoff factor.

(a) For  0.20 m C12H22O11, m = 0.20 molKg-1 and i = 1

Hence DeltaTf = i x Kf x m = 1x(- 1.86 DegCKgmol-1)x (0.20 molKg-1) =   - 0.372 DegC

=> Tf - 0 = -0.372 DegC

=> Tf = - 0.372 DegC

(b) For 0.15 m NaCl, m = 0.15 molKg-1 and i = 2

Hence DeltaTf = i x Kf x m = 2x(- 1.86 DegCKgmol-1)x (0.15 molKg-1) =   - 0.558 DegC

=> Tf - 0 = -0.558 DegC

=> Tf = - 0.558 DegC

(c):  For 0.10 m HCl

m = 0.10 molKg-1 and i = 2 (nearly equals to 2)

Hence DeltaTf = i x Kf x m = 2x(- 1.86 DegCKgmol-1)x (0.10 molKg-1) =   - 0.372 DegC

=> Tf - 0 = -0.372 DegC

=> Tf = - 0.372 DegC

(d): For 0.050 m CaCl2

m = 0.050 molKg-1 and i = 3 (nearly equals to 3)

Hence DeltaTf = i x Kf x m = 3x(- 1.86 DegCKgmol-1)x (0.050 molKg-1) =   - 0.279 DegC

=> Tf - 0 = -0.279 DegC

(e): For 0.050 m CH3COOH

m = 0.050 molKg-1 and i = 1 (nearly equals to 1)

Hence DeltaTf = i x Kf x m = 1x(- 1.86 DegCKgmol-1)x (0.050 molKg-1) =   - 0.093DegC

=> Tf - 0 = -0.093 DegC

Hence (b)  0.15 m NaCl has the lowest freezing point (answer)

(4): Let the molecular mass of the powder be 'M'

Mass of the white powder dissolved, m = 82 mg = 82 mg x (1g / 1000 mg) = 0.082 g

Moles of the white powder, n = mass / mlecular mass = 0.082 / M mol

volume of solvent ethanol, V = 1.50 mL

Density of ethanol = d = 0.789 g/cm3 = 0.789 g/mL

Hence mass of ethanol = Vxd = 1.50mL x 0.789 g/mL = 1.1835 g = (1.1835 g)x (1 Kg / 1000g) = 0.0011835 Kg

Molality, m = (Moles of solute) / mass of solvent (in Kg) = (0.082 / M mol) / (0.0011835 Kg) = (69.29 / M) m

Kf of ethenol, Kf = 1.99 C/m

All the given compounds have Van't hoff equals to 1.

Hence DeltaTf for the given solution can be calculated as

DeltaTf = -1xKfxm = - (1.99 C/m) x (69.29 / M) m

=>[ - 115.5 - (-114.6)] DegC = -137.89 / M

=> - 0.9 DegC = -137.89 / M

=> M = 153.2 g/mol

Among the given compounds norfenefrine (C8H11NO2) has a molecular mass equals to 153.2 g/mol. Hence the white powder is norfenefrine (C8H11NO2) (answer)

It needs more information to answer 1 and 2.

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