solution chemistry + concentration
Define a solution and state whether dissolving is a physical or chemical change.
A solution is a homogeneous mixture of two or more pure substances. Dissolving is a physical change because the solute can be recovered by evaporating the solvent.
List three colligative properties.
Vapor pressure lowering, boiling point elevation, freezing point depression (osmotic pressure is also acceptable).
Define an exothermic reaction in terms of ΔH and heat flow.
An exothermic reaction releases heat to the surroundings, so ΔH < 0.
True or False: ΔH is a state function. Explain briefly.
True. ΔH depends only on the initial and final states, not the path taken.
Write the First Law of Thermodynamics as an equation and in words.
ΔE = q + w. Energy is neither created nor destroyed; it is conserved and can be transferred as heat or work.
Calculate the molarity of a solution made by dissolving 0.75 moles of NaCl in enough water to make 1.5 L of solution.
M = mol solute / L solution = 0.75 mol / 1.5 L = 0.50 M
Calculate the boiling point elevation of a 0.50 m aqueous glucose solution (Kb water = 0.51 °C/m).
ΔTb = Kb × m = 0.51 × 0.50 = 0.255°C
Calculate the heat required to raise the temperature of 50.0 g of water from 20.0°C to 35.0°C. (Specific heat of water = 4.184 J/g·K)
q = m × s × ΔT = 50.0 × 4.184 × 15.0 = 3138 J = 3.14 kJ
Calculate ΔH° for the reaction: 2H₂(g) + O₂(g) → 2H₂O(l) using standard enthalpies of formation: ΔH°f H₂O(l) = –286 kJ/mol.
ΔH° = Σ n ΔH°f(products) – Σ m ΔH°f(reactants)
= 2(–286) – [2(0) + 1(0)] = –572 kJ
A gas expands from 2.0 L to 5.0 L against a constant external pressure of 1.5 atm. Calculate the work done by the gas (1 L·atm = 101.3 J).
W = –PΔV = –1.5 atm × (5.0 – 2.0) L = –4.5 L·atm
= –4.5 × 101.3 = –456 J
A solution contains 5.00 g of glucose (C₆H₁₂O₆) in 100.0 g of water. Calculate the mass percentage and the molality of glucose.
Mass % = (5.00 / 105.0) × 100 = 4.76%
Molality = (5.00 g / 180.16 g/mol) / 0.1000 kg = 0.277 m
What is the freezing point of a 0.75 m aqueous NaCl solution? (Kf water = 1.86 °C/m, assume complete dissociation, i = 2)
ΔTf = i × Kf × m = 2 × 1.86 × 0.75 = 2.79°C; Freezing point = 0.00 – 2.79 = –2.79°C
In a constant-pressure calorimeter, 0.500 g of Mg reacts with excess HCl, raising the temperature of 100.0 g of water by 5.30°C. Calculate the heat released by the reaction per gram of Mg.
q_water = 100.0 × 4.184 × 5.30 = 2218 J
Heat per gram Mg = 2218 J / 0.500 g = 4436 J/g = 4.44 kJ/g
Given the following reactions:
(1) C(s) + O₂(g) → CO₂(g) ΔH = –394 kJ
(2) H₂(g) + ½O₂(g) → H₂O(l) ΔH = –286 kJ
(3) C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l) ΔH = –1367 kJ
Find ΔH°f for C₂H₅OH(l).
Target: 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)
ΔH°f = 2(–394) + 3(–286) – (–1367) = –788 –858 +1367 = –279 kJ/mol
A system absorbs 250 J of heat and has 100 J of work done on it. Calculate ΔE.
ΔE = q + w = (+250) + (+100) = +350 J
Using Henry’s Law, calculate the solubility of a gas in water if the partial pressure above the liquid is 2.50 atm and k = 0.034 M/atm.
S = k × P = 0.034 M/atm × 2.50 atm = 0.085 M
A solution contains 10.0 g of an unknown non-electrolyte dissolved in 500.0 g of water. The freezing point is –0.372°C. Find the molar mass of the solute. (Kf = 1.86)
ΔTf = 0.372 = 1.86 × m → m = 0.200 mol/kg
Moles solute = 0.200 mol/kg × 0.5000 kg = 0.100 mol
Molar mass = 10.0 g / 0.100 mol = 100 g/mol
A bomb calorimeter has a heat capacity of 2.50 kJ/°C. A 0.200 g sample of a fuel raises the temperature from 22.0°C to 26.5°C. Calculate the heat released per mole if the fuel’s molar mass is 100 g/mol.
q_cal = C × ΔT = 2.50 × (4.5) = 11.25 kJ
Moles fuel = 0.200 / 100 = 0.00200 mol
ΔE per mol = 11.25 kJ / 0.00200 mol = 5625 kJ/mol
Estimate ΔH for the reaction: H–H + Cl–Cl → 2H–Cl using bond enthalpies (H–H = 436, Cl–Cl = 243, H–Cl = 431 kJ/mol).
ΔH = Σ bonds broken – Σ bonds formed = (436 + 243) – (2 × 431) = 679 – 862 = –183 kJ
In an open vessel (constant pressure), a reaction releases 500 J of heat and does 200 J of expansion work. Calculate ΔE and ΔH.
ΔH = q_p = –500 J
ΔE = q + w = (–500) + (–200) = –700 J
A solution is prepared by mixing 25.0 g of ethanol (C₂H₅OH, molar mass 46.07 g/mol) and 75.0 g of water. Calculate the mole fraction of ethanol and the vapor pressure of the solution at 25°C (pure water vapor pressure = 23.8 torr). Assume ethanol is nonvolatile.
Moles ethanol = 25.0 / 46.07 = 0.543 mol
Moles water = 75.0 / 18.02 = 4.16 mol
X_water = 4.16 / (4.16 + 0.543) = 0.885
P_solution = X_water × P°_water = 0.885 × 23.8 = 21.1 torr
At 25°C, the osmotic pressure of a solution containing 2.50 g of a protein in 100.0 mL of solution is 0.0180 atm. Calculate the molar mass of the protein. (R = 0.08206 L·atm/(mol·K))
π = MRT → 0.0180 = M × 0.08206 × 298 → M = 0.000736 mol/L
Moles protein in 0.100 L = 0.000736 × 0.100 = 7.36 × 10⁻⁵ mol
Molar mass = 2.50 g / 7.36×10⁻⁵ mol = 33,967 g/mol ≈ 3.40 × 10⁴ g/mol
In a calorimetry experiment, initial temperature = 20.0°C, final measured temperature = 28.0°C, but the cooling curve extrapolation shows the true final temperature should be 29.5°C. If the calorimeter contains 150.0 g water (specific heat 4.184), calculate the percentage of heat leaked to the surroundings.
Heat without leakage = 150 × 4.184 × (29.5 – 20.0) = 150 × 4.184 × 9.5 = 5962 J
Heat measured = 150 × 4.184 × (28.0 – 20.0) = 150 × 4.184 × 8.0 = 5021 J
Heat leaked = 5962 – 5021 = 941 J
% leaked = (941 / 5962) × 100 = 15.8%
The average bond enthalpy of the O–H bond in water is 463 kJ/mol, but the actual bond enthalpy for the first O–H bond in H₂O is 497 kJ/mol, and for the second is 424 kJ/mol. Explain why these differ, then calculate the average from these two values and compare to the given average.
After the first H is removed, the remaining OH⁻ has stronger attraction between O and H (more electron density per bond), so the first bond is harder to break than the second.
Average = (497 + 424) / 2 = 460.5 kJ/mol, which matches the given average of 463 kJ/mol (small rounding difference).
A gas in a cylinder with a piston absorbs 800 J of heat while the surroundings do 200 J of compression work. Then, the gas expands against a constant pressure of 2.0 atm from 1.0 L to 4.0 L while losing 100 J of heat. Calculate the net ΔE for the two-step process.
Step 1: q = +800 J, w = +200 J (work done on system) → ΔE₁ = +1000 J
Step 2: q = –100 J, w = –PΔV = –2.0 × (4.0 – 1.0) = –6.0 L·atm = –6.0 × 101.3 = –608 J → ΔE₂ = –100 – 608 = –708 J
Net ΔE = 1000 + (–708) = +292 J