Gas Laws, Ideal gas Eq
A gas at 300 K and 100 kPa occupies 2.00 dm³. When the conditions change to 400 K and 150 kPa, this is the new volume.
Correct response: What is 1.78 dm³?
Working: P₁V₁/T₁ = P₂V₂/T₂ (100 × 2.00) / 300 = (150 × V₂) / 400 V₂ = (100 × 2.00 × 400) / (300 × 150) V₂ = 1.78 dm³
What is a mole?
A mole is a unit used to count a large amount particles like atoms, compounds, or ions.
One mole of substance equals 6.022 × 10²³ particles, which is known as Avogadro's number.
This formula shows the simplest whole-number ratio of atoms.
What is c = n / V?
This is the formula used to calculate concentration when amount in moles and volume in dm³ are known.
A gas sample has a volume of 500 cm³ at 25°C and 100 kPa. This is the temperature in Kelvin.
Correct response: What is 298 K?
Working: T(K) = 25 + 273 = 298 K
Find the mass in grams of 2.00 x 1023 molecules of F2.
12.7 g
Give the empirical formula of acetic acid (C2H4O2).
CH2O.
A solution contains 0.250 mol of sodium chloride dissolved in 0.500 dm³ of solution. This is the concentration of the solution.
What is 0.500 mol dm⁻³?
Working: c = 0.250 / 0.500 = 0.500 mol dm⁻³
Clue: A gas occupies 5.00 dm³ at 298 K and 101 kPa. When cooled to 273 K at constant pressure, this is the new volume.
Correct response: What is 4.58 dm³?
Working: V₁/T₁ = V₂/T₂ 5.00 / 298 = V₂ / 273 V₂ = (5.00 × 273) / 298 V₂ = 4.58 dm³
How many particles are in 3 mol glucose?
3 × 6.022×10²³ = 1.8066×10²⁴ molecules of glucose
The empirical formula is NH₂. The molar mass is 32.0 g mol⁻¹. This is the molecular formula.
Correct response: What is N₂H₄?
Working: Molar mass of NH₂ = 14.0 + 2(1.0) = 16.0 g mol⁻¹ n = 32.0 / 16.0 = 2 Molecular formula = (NH₂)₂ = N₂H₄
This is the dilution formula used when concentration and volume change but moles remain constant.
Correct response: What is c₁V₁ = c₂V₂?
0.500 mol of an ideal gas is at 101 kPa and 298 K. This is the volume in dm³.
Correct response: What is 12.0 dm³?
Working: PV = nRT 101 × V = 0.500 × 8.31 × 298 V = (0.500 × 8.31 × 298) / 101 V = 12.0 dm³
One mole of an element has a mass equal to this, in g/mol.
Atomic Mass or Molecular Mass
A compound has empirical formula CH₂O and molar mass 180 g/mol, what is its molecular formula?
C₆H₁₂O₆
25.0 cm³ of 1.00 mol dm⁻³ hydrochloric acid is diluted to 250 cm³. This is the new concentration
What is 0.100 mol dm⁻³?
Working: c₁V₁ = c₂V₂ 1.00 × 25.0 = c₂ × 250 c₂ = 0.100 mol dm⁻³
: A compound has the empirical formula C₂H₆O and a molar mass of 92.0 g mol⁻¹. A 1.84 g sample of this compound is vaporized and occupies 1.00 dm³ at 373 K. This is the pressure in kPa.
Correct response: What is 46.0 kPa?
Working: Step 1: Find molecular formula Molar mass of C₂H₆O = 2(12.0) + 6(1.0) + 16.0 = 46.0 g mol⁻¹ n = 92.0 / 46.0 = 2 Molecular formula = (C₂H₆O)₂ = C₄H₁₂O₂
Step 2: Find moles of gas n = 1.84 / 92.0 = 0.0200 mol
Step 3: Use ideal gas equation PV = nRT P × 1.00 = 0.0200 × 8.31 × 373 P = 46.0 kPa
What is the molar mass of H3PO4?
98 g/mol
A compound contains 40.0% C, 6.7% H, and 53.3% O by mass. The molar mass is 60.0 g mol⁻¹. This is the molecular formula.
Correct response: What is CH₂O?
Working: Empirical formula (from Category 1, 200 points) = CH₂O Molar mass of CH₂O = 30.0 g mol⁻¹ n = 60.0 / 30.0 = 2 Molecular formula = (CH₂O)₂ = C₂H₄O₂
A student dilutes 20.0 cm³ of sodium hydroxide solution to 250.0 cm³. The final concentration is 0.0800 mol dm⁻³. This was the original concentration.
What is 1.00 mol dm⁻³?
Working: c₁V₁ = c₂V₂ c₁ × 20.0 = 0.0800 × 250.0 c₁ = 1.00 mol dm⁻³