Question

9) An aqueous solution of HCl is 0.500 M. Its density is 1.20 g/mL. Calculate the molality of HCl in this solution.

a) 0.417 m

b) 0.424 m

c) 0.430 m

d) 0.500 m

e) 0.600 m

10) The normal freezing point of benzene (C₆H₆) is 5.48°C. When 2.00 g of an unknown covalent compound is dissolved in 100.0 g of benzene, the freezing point of the resulting solution is 5.08°C. What is the molar mass of the unknown compound? For benzene, K_f = 5.12°C/m.

a) 26.0 g/mol

b) 78.0 g/mol

c) 128 g/mol

d) 256 g/mol

e) 1280 g/mol

Hint: For a covalent compound, van't Hoff factor i = 1.

12) Place the following aqueous solutions in order of increasing osmotic pressure at 298 K.

            I. 0.10 M glucose, C₆H₁₂O₆                II. 0.10 M Ca(NO₃)₂               III. 0.10 M NaCl

a) III < I < II

b) II < III < I

c) I < II < III

d) II < I < III

e) I < III < II

Answer 1

9) An aqueous solution of HCl is 0.500 M. Its density is 1.20 g/mL. Calculate the molality of HCl in this solution.

molality = mol of HCl / kg of water

assume a basis of 1000 mL of solution

mass od solution = D*V = 1.20*1000 = 1200 g

mol of HCl = 0.5

mass = mol*MW = 0.5*36.65 = 18.325 g

mass of solvent = 1200-18.325 = 1181.675 g of water = 1.18 kg

molality = mol/kg = 0.5/1.18

molal = 0.4237

10) The normal freezing point of benzene (C6H6) is 5.48°C. When 2.00 g of an unknown covalent compound is dissolved in 100.0 g of benzene, the freezing point of the resulting solution is 5.08°C. What is the molar mass of the unknown compound? For benzene, Kf = 5.12°C/m.

c= mol/kg solvent = (2/MW) / (0.1) = 20/MW

dTf = 5.08-5.48 = -0.40

-0.40/-5.12 = molal

20 /0.078= MW

MW = 256 g/mol

12) Place the following aqueous solutions in order of increasing osmotic pressure at 298 K.

Posm = i*MR*T

if R, T constant, then i*M is important in this case

note that M is the same for all, then compare "i2 only

glucose i = 1

Nacl, i = 2

Ca(NO3)2 = i = 3

then

I < III < II