CHEM 103 MODULE 6 PORTAGE LEARNING CHEMISTRY EXAM

CHEM 103 MODULE 6 PORTAGE LEARNING CHEMISTRY EXAM 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 (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. CHEM 103 CHEM 103 (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. (i) Fluid (s) Vapor pressure (j) Ionization (t) Weak electrolyte What rule is used to predict the solubility of materials? 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 Explain how and why the presence of a solute affects the boiling point of a solvent. dissolving nonpolar substances. 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 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. Molality = (gsolute / MW) / (gsolvent / 1000) Show the calculation of the molarity of a solution made by dissolving 12.6 grams of NaNO3 to make 200 ml of solution Molality = (12.6 / 85.0) / (200 / 1000) = 0.741 m Molarity = (gsolute / MW) / (mlsolution / 1000) Show the calculation of the mass of NaNO3 needed to make 450 ml of a 0.356 M solution Molarity = (12.6 / 85.0) / (200 / 1000) = 0.741 M . 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 Molessolute = gsolute / MW Molessolute = 24.6 g / 85.0 = 0.2894 mol Molarity = moles / (mL /1000) CHEM 103 CHEM 103 0.987 = 0.2894 / (mL / 1000) mL / 1000 = 0.2894 / 0.987 = 0.2932 mL = 0.2932 x 1000 = 293 mL . molality = (gsolute / MW) / (gsolvent / 1000) molality = (12.6 / 46.07) / (200 / 1000) = 1.37 m ∆tf = Kf x m = 1.86 x 1.37 = 2.55 C FPsolution = FPsolvent - ∆tf = 0 C – 2.55 = - 2.55 C 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 0C. 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 = 2.76 C FPsolution = FPsolvent - ∆tf = 0 C – 2.76 = - 2.76 C 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 0 C. ∆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 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 -1.20 C. Kf for water is 1.86 and the FP of pure water is 0 C. AlCl3 → Al+3 + 3 Cl- ∆tf = 1.86 x 0.1 x 4 = 2nd lowest FP Ca3(PO4)2 → 3 Ca+2 + 2 PO4-3 ∆tf = 1.86 x 0.1 x 5 = lowest FP KCl → K+ + Cl- ∆tf = 1.86 x 0.1 x 2 = highest FP CaCl2 → Ca+2 + 2 Cl- ∆tf = 1.86 x 0.1 x 3 = 3rd lowest FP FP: Ca3(PO4)2 < AlCl3 < CaCl2 < KCl 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 Milk is an emulsion (liquid in liquid), whipped cream is a foam (gas in liquid) and fog is an aerosol (liquid in gas). Categorize (1) milk, (2) whipped cream and (3) fog as emulsion, foam or aerosol. CHEM 103 CHEM 103 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. Explain how and why the presence of a solute affects the boiling point of a solvent. GaCl3 → Ga+3 + 3 Cl- ∆tf = 1.86 x 0.1 x 4 = 3rd lowest FP Al2(SO4)3 → 2 Al+3 + 3 SO4-3 ∆tf = 1.86 x 0.1 x 5 = lowest FP NaI → Na+ + I- ∆tf = 1.86 x 0.1 x 2 = highest FP MgCl2 → Mg+2 + 2 Cl- ∆tf = 1.86 x 0.1 x 3 = 2nd lowest FP FP: Al2(SO4)3 < GaCl3 < MgCl2 < NaI Rank and explain how the freezing point of 0.100 m solutions of the following ionic electrolytes compare, List from lowest freezing point to highest freezing point. GaCl3, Al2(SO4)3, NaI, MgCl2 Mass % = (gsolute / gsolute + gsolvent) x 100% Mass % = (15.7 / 15.7 + 350) x 100 = 4.29% Show the calculation of the mass percent solute in a solution of 15.7 grams of Ca3(PO4)2 in 350 grams of water. Report your answer to 3 significant figures. molality = (gsolute / MW) / (gsolvent / 1000) molality = (24.5 / 120.104) / (400 / 1000) = 0.510 m Show the calculation of the molality of a solution made by dissolving 24.5 grams of C4H8O4 in 400 grams of water. Report your answer to 3 significant figures. Molarity = (gsolute / MW) / (mlsolvent / 1000) Molarity = (27.5 / 310.18) / (350 / 1000) = 0.253 M Show the calculation of the molarity of a solution made by dissolving 27.5 grams of Ca3(PO4)2 to make 350 ml of solution. Report your answer to 3 significant figures. Molarity = (moles) / (mlsolvent / 1000) 0.200 = (moles) / (250 / 1000) Moles = 0.200 x 0.250 = 0.0500 Moles = (gsolute / MW) 0.0500 = (gsolute / 375.41) gsolute = 0.0500 x 375.41 = 18.8 g Show the calculation of the mass of Ba(MnO4)2 needed to make 250 ml of a 0.200 M solution. Report your answer to 3 significant figures. molessolute = gsolute / MW molessolute = 37.3 g / 164.10 = 0.2273 mol Molarity = moles / (mL /1000) 0.756 = 0.2273 / (mL / 1000) mL / 1000 = 0.2273 / 0.756 = 0.3007 mL = 0.3007 x 1000 = 301 mL Show the calculation of the volume of 0.756 M solution which can be prepared using 37.3 grams of Ca(NO3)2. molality = (gsolute / MW) / (gsolvent / 1000) molality = (32.6 / 150.13) / (200 / 1000) = 1.08