Module 6 Problem Set Quiz
Module 6 Problem Set
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Define or describe:
(a) Amorphous solid (k) Molality
(b) Colligative property (l) Molarity
(c) Colloid (m) Nonelectrolyte
(d) Condensation (n) Phase
(e) Critical point (o) Strong electrolyte
(f) Critical pressure (p) Sublimation
(g) Crystalline solid (q) Surface tension
(h) Electrolyte (r) Triple point
(i) Fluid (s) Vapor pressure
(j) Ionization (t) Weak electrolyte
Correct Answer
(a) Amorphous solids are ones in which the particles are arranged in a random fashion.
(b) Colligative properties are physical properties of solvents that are dependent on the concentration of the solute present and the identity of the solvent but not on the identity of the solute. They include vapor pressure, freezing point and boiling point.
(c) Colloids are heterogeneous mixtures which appear to homogeneous one-phase mixtures but are actually composed of particles too small to be seen with the naked eye.
(d) Condensation is the process of gas being converted to liquid.
(e) Critical point is the temperature above which a substance cannot exist in the liquid phase.
(f) Critical pressure is the lowest pressure required for the substance to exist as a liquid at the critical point.
(g) Crystalline solids are ones in which the particles are arranged in one of several different orderly, repeating, geometric patterns.
(h) Electrolytes are ionic or very polar compounds which dissolve to form solutions of ions which conduct an electric current.
(i) Fluids are substances like liquids and gases which have no fixed shape and so they flow.
(j) Ionization is the splitting of molecules to form ions.
(k) Molality is a term which expresses concentration in moles of solute present per kilogram of solvent.
(l) Molarity is a term which expresses concentration in moles of solute present per liter of solution.
(m) Nonelectrolytes are compounds which dissolve to form solutions of molecules which do not conduct an electric current.
(n) Phase is any state of matter such as solid, liquid or gas.
(o) Strong electrolytes are solutes that ionize completely.
(p) Sublimation is the conversion of a solid directly to the gas state.
(q) Surface tension is the force that causes a liquid which is in contact with a gas like air to assume a shape that has the least amount of surface area causing the surface to act like a thin elastic sheet.
(r) Triple point is the temperature and pressure at which the solid, liquid and gas phases can coexist.
(s) Vapor pressure is pressure exerted by vapor molecules above a liquid.
(t) Weak electrolytes are solutes that only partially ionize.
What rule is used to predict the solubility of materials?
Correct Answer
The "like dissolves like" rule is used to predict the solubility of materials with polar solvents only dissolving polar (and ionic) substances and nonpolar solvents only dissolving nonpolar substances.
Explain how and why the presence of a solute affects the boiling point of a solvent.
Correct Answer
The presence of a solute raises the boiling point of a solvent by lowering the vapor pressure of the solvent. With this lower vapor pressure, more heat (a higher boiling point) is required to raise the vapor pressure to atmospheric pressure.
Calculate the mass percent solute in a solution of 12.6 grams of NaNO3 in 450 grams of water.
Correct Answer
Mass % = (gsolute / gsolute + gsolvent) x 100%
Mass % = (12.6 / 12.6 + 450) x 100 = 2.72%
Show the calculation of the molality of a solution made by dissolving 12.6 grams of NaNO3 in 200 grams of water.
Correct Answer
molality = (gsolute / MW) / (gsolvent / 1000)
molality = (12.6 / 85.0) / (200 / 1000) = 0.741 m
Show the calculation of the molarity of a solution made by dissolving 12.6 grams of NaNO3 to make 200 ml of solution.
Correct Answer
Molar mass of NaNO3=22.99+14.01+3(16.00)=85.00 g/mol
# mole of NaNO3=12.6 g/85.00 g/mol= 0.148 mol
Molarity = (gsolute / MW) / (mlsolution / 1000)
Molarity =0.148 mol/(200 mL*1 L/1000 mL)=0.741 M
Show the calculation of the mass of NaNO3 needed to make 450 ml of a 0.356 M solution.
Correct Answer
Molarity = (moles) / (mlsolution / 1000)
0.356 = (moles) / (450 / 1000)
Moles = 0.356 x 0.450 = 0.1602
Moles = (gsolute / MW)
0.1602 = (gsolute / 85.0)
gsolute = 0.1602 x 85.0 = 13.6 g
Show the calculation of the volume of 0.987 M solution which can be prepared using 24.6 grams of NaNO3.
Correct Answer
Molar mass of NaNO3=22.99+14.01+3(16.00)=85.00 g/mol
molessolute = gsolute / MW
molessolute = 24.6 g / 85.0 = 0.2894 mol
Molarity = moles / (mL /1000)
0.987 = 0.2894 / (mL / 1000)
mL / 1000 = 0.2894 / 0.987 = 0.2932
mL = 0.2932 x 1000 = 293 mL
Show the calculation of the freezing point of a solution made by dissolving 12.6 grams of the nonelectrolyte C2H5OH in 200 grams of water. Kf for water is 1.86, FP of pure water is 0oC.
Correct Answer
Molar mass of C2H5OH=2(12.01)+6(1.008)+16.00=46.07 g/mol
# moles of C2H5OH=12.6 g/46.07 g/mol= 0.273 mol
Mass of solution =12.6+200=212.6 g
Molality= 0.273 mol/(212.6 g*1kg/1000 g)=1.37 m
Show the calculation of the freezing point of a solution made by dissolving 12.6 grams of the electrolyte NaNO3 in 200 grams of water. Kf for water is 1.86 and the FP of pure water is 0oC.
Correct Answer
molality = (gsolute / MW) / (gsolvent / 1000)
molality = (12.6 / 85.0) / (200 / 1000) = 0.741 m
NaNO3 → Na+ + NO3- (2 moles of particles)
∆tf = Kf x m x 2 = 1.86 x 0.741 x 2 = 2.76oC
FPsolution = FPsolvent - ∆tf = 0oC – 2.76 = - 2.76oC
Show the calculation of the molar mass (molecular weight) of a solute if a solution of 5.8 grams of the solute in 100 grams of water has a freezing point of -1.20oC. Kf for water is 1.86 and the FP of pure water is 0oC.
Correct Answer
∆tf = Kf x m
molality = ∆tf / Kf = 1.20 / 1.86 = 0.645 m
molality = (gsolute / MW) / (gsolvent / 1000)
0.645 = (moles) / (100 / 1000)
Moles = 0.645 x 0.100 = 0.0645
0.0645 = (5.8 / MW)
MW = 5.8 / 0.0645 = 89.9
How would the FP of 0.100 m solutions of the following ionic electrolytes compare? Rank from lowest FP to highest FP.
AlCl3, Ca3(PO4)2, KCl, CaCl2
Correct Answer
