9) An aqueous solution of HCl is 0.500 M . Its density is 1.20 g/mL. Calculate t
ID: 586864 • Letter: 9
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, Kf = 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
Explanation / Answer
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
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