How many moles of ions are contained in 1 91 L of a 1 59 M solution of KCl

# How many moles of ions are contained in 1 91 l of a 1

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How many moles of ions are contained in 1 . 91 L of a 1 . 59 M solution of KCl? Correct answer: 6 . 0738 mol. Explanation: solute = KCl n = ? KCl(s) H 2 O ----→ K + (aq) + Cl (aq) Ions produced = 1 mol K + + 1 mol Cl = 2 mol 023(part2of2)10.0points How many moles of ions are contained in 1 . 91 L of a 1 . 59 M solution of Mg(NO 3 ) 2 ? Correct answer: 9 . 1107 mol. Explanation: solute = Mg(NO 3 ) 2 n = ? Mg(NO 3 ) 2 (s) H 2 O ----→ Mg 2+ (aq) + 2NO 3 (aq) Ions produced = 1 mol Mg 2+ + 2 mol NO 3 = 3 mol 024 10.0points Raising the temperature of a sample of water will typically (decrease/increase) the solubil- ity of (all/some) dissolved gases. 1. decrease, some 2. increase, some 3. increase, all 4. decrease, all correct Explanation: Because the dissolution of most all gases is an exothermic process, Le Chatelier’s prin- ciple suggests that all gases will become less soluble as the temperature of the solvent is raised. 025 10.0points To increase the solubility of a gas in water, you should 1. increase the pressure of the gas. correct 2. increase the temperature of the water. 3. increase the surface area of the particles. 4. decrease the pressure of the gas. Explanation: 026 10.0points Theoretically, it would be harder to dis- solve (NaCl/Al 2 S 3 ) in water because the (higher/lower) the charge density, the lower the solubility. 1. Al 2 S 3 , higher correct 2. NaCl, lower 3. NaCl, higher 4. Al 2 S 3 , lower Explanation:
nanni (arn437) – HW 02 Solutions – vanden bout – (51165) 7 Al 2 S 3 has a higher charge density (Al 3+ , S 2 ) than NaCl (Na + , Cl ). High charge density corresponds to high lattice energy and thus much more endothermic (i.e. less spon- taneous) dissolution. 027 10.0points The freezing point of seawater is about - 1.85 C. If seawater is an aqueous solution of sodium chloride, calculate the molality of seawater. The k f for water is 1.86 K/ m . 1. 3.44 m 2. 0.995 m 3. 3.70 m 4. 1.99 m 5. 0.497 m correct Explanation: 028 10.0points If acetic acid has a pure vapor pressure of 20 torr at 30 C and acetaldehye has a pure vapor pressure of 1000 torr at 30 C, a mix- ture of 2 moles of acetic acid and 8 moles of acetaldehyde would have what total vapor pressure? 1. 804 torr correct 2. 216 torr 3. 204 torr 4. 510 torr 5. 816 torr Explanation: P total = χ a · P a + χ b · P b + ... P total = 0 . 2 · 20 torr + 0 . 8 · 1000 torr P total = 804 torr 029 10.0points The vapor pressure of a volatile component in a solution decreases as its mole fraction decreases. This is known as 1. Bragg’s Law. 2. Raoult’s Law. correct 3. LeChatelier’s Principle. 4. Henry’s Law. Explanation: According to Raoult’s Law, the vapor pres- sure of the volatile component of a solution equals its mole fraction in the solution times the vapor pressure of the pure solvent. P = X P 0 030 10.0points What is the vapor pressure of the solvent in an aqueous solution at 25 C of 0 . 65 m urea (CO(NH 2 ) 2 ), a nonelectrolyte? The vapor pressure of water at 25 C is 23 . 76 Torr.

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• Fall '07
• Holcombe
• pH, Solubility, Vapor pressure, Sodium chloride, mol, Vanden Bout