Chp 13 Quiz Flashcards
(10 cards)
Identify the compounds that are soluble in both water and hexane
- 1-propanol and 1-pentanol
- 1-butanol and 1-pentanol
- Ethanol and 1-propanol
- Ethanol and 1-butanol
- Methanol and 1-pentanol
3
-Explanation: like dissolves like
> Polar compounds soluble in water; non-polar compounds soluble in hexane
> Alcohol: has both non-polar alkyl gp + polar OH gp
> Solubility depends on size of alkyl gp
> Increase in size > More significant fraction of alcohol molecule
> Less soluble in water
> @Rm temp: gp w/ 4 carbons (<4C=soluble in water; >4C=insoluble in water)
> Becoz OH gp can drag abt 3-4 carbons into solution in water
Determine Change in H_solute for KBr if
Change in H_solution (KBr) = +19.9 kJ/mol
Change in H_hydration (KBr) = -670. kJ/mol
+690. kJ/mol
- Explanation: Change in H_solution = Change in H_solute + Change in H_hydration
> 19.9 = ? - 670
Calculate the mass of oxygen (in mg) dissolved in a 5.00L bucket of water exposed to a pressure of 1.13 atm of air. Assume the mole fraction of oxygen in air to be 0.21 given that
k_H for O2 is 1.3 * 10^-3 M/atm at this temperature
Step 1: Partial pressure of oxygen
= mole fraction of oxygen * pressure of air
= 0.21 * 1.13 = 0.2373 atm
Step 2: Concentration of oxygen
= Henry’s law constant * partial pressure of oxygen
=1.3 * 10^-3 * 0.2373 = 3.0849 * 10^-4 M
Step 3: Moles of oxygen in 5.00L
= volume * concentration
= 5.00 * 3.0849 * 10^-4 = 1.54245 * 10^-3 mol
Step 4: Mass of oxygen = moles * molar mass of oxygen
= 1.54245 * 10^-3 * 32.00 = 49.4 mg
A solution of LiCl in water has X_LiCl = 0.0900
What is the molality?
Mole fraction of LiCl = 0.0900
> Mole fraction of H2O = 0.91
> Mole of LiCl =
At 20C, a 0.376M aqueous solution of ammonium chloride has a density of 1.0045 g/mL.
What is the mass % of ammonium chloride in the solution?
The formula weight of NH4Cl is 53.50 g/mL
Step 1: Mole of NH4Cl = molarity * volume = 0.376 M * 1L
= 0.376 mol
Step 2: Mass of NH4Cl = moles * molar mass = 0.376 mol * 53.50 g/mol
= 20.116 g
Step 3: Mass of solution = density * volume = 1.0045 g/mL * 1000mL
= 1004.5 g
Step 4: Mass % of NH4Cl = (mass of NH4Cl / 10004.5 g) * 100%
= 20.116 * 100 / 1004.5 = 2.00%
The osmotic pressure of a solution formed by dissolving 50.0 mg of aspirin (C9H8O4) in enough water to make 0.250 L of solution at 25C is _________ atm
Pi * V = n * S * T
Where Pi = osmotic pressure of a solution, V = volume = 0.25L,
n = moles = mass / molar mass = 0.05 / 180 = 2.778 * 10^-4 mol,
S = 0.08206 atmL (OR molK), T = 25 + 273 = 298 K
> Pi = 0.0272 atm
Calculate the freezing point of a solution containing 10 g of KCl and 1100.0 g of water.
The molal freezing point depression constant (K_f) for water is 1.86 C/m
Step 1: Molar mass of KCl = 74.5513 g/mol
Step 2: Molality (m) = mass of KCl * 1000 / (molar mass of KCl + mass of water) = 0.1219 m
Step 3: Freezing point Change in T_R = i * k_f * m
i = Van’t Hoff’s factor for KCl = 2
K_f = Molal freezing depression in freezing point constant = 1.86
> Freezing point Change in T_R = 2 * 1.86 * 0.1219 = 0.454 C
Step 4: Freezing point of solution = 0 C - 0.454 C = -0.454 C
Choose the aqueous solution below with the lowest freezing point.
These are all solutions of nonvolatile solutes and you should assume ideal van’t Hoff factors where applicable.
A. 0.075 m LiCN
B. 0.075 m KNO2
C. 0.075 m (NH4)3PO4
D. 0.075 m NaIO4
E. 0.075 m LiCl
C
> 0.075m * 4
At a given temperature the vapor pressures of benzene and toluene are 183 mmHg and 59.2 mmHg, respectively.
Calculate the total vapor pressure over a solution of benzene and toluene with X_benzene = 0.580
P_total = P_benzene + P_toluene
= X_benzene * P_pure benzene + X_toluene * P_pure toluene
= 0.580 * 183 + 0.420 * 59.2 = 131 mmHg
Determine the boiling point of a solution that contains 78.8 g of naphthalene (C10H8, molar mass = 128.16 g/mol) dissolved in 722 mL of benzene (d = 0.877 g/mL).
Pure benzene has a boiling point of 80.1 C and a boiling point elevation constant of 2.53 C/m
Step 1: Moles of naphthalene = mass / molar mass = 78.8 / 128.16
= 0.61486 mol
Step 2: Mass of benzene = volume * density = 722 * 0.877 = 633.194 g
= 0.633194 kg
Step 3: Molality = moles of naphthalene / mass of benzene
= 0.61486 / 0.633194 = 0.97104 m
Step 4: Change in T_b = K_b * molality
= 2.53 * 0.97104 = 2.4567 C
*Step 5: BP of solution = BP of benzene + Change in T_b
= 80.1 C + 2.4567 C = 82.6 C
