The diagram shows the apparatus used to find the relative molecular mass of a volatile liquid. When 0.10 g of a volatile liquid is injected into the syringe, all of the volatile liquid evaporates and the volume increases by 85 cm³. The heater maintains a temperature of 400K and the experiment is carried out at a pressure of 101 300Pa.
If the vapour of the volatile liquid behaves as an ideal gas, which expression can be used to calculate the relative molecular mass of the liquid?
▶️ Answer/Explanation
Ans: C
Using the ideal gas equation \( PV = nRT \), where \( n = \frac{m}{M} \), we substitute the given values: mass \( m = 0.10 \, \text{g} \), volume \( V = 85 \times 10^{-6} \, \text{m}^3 \), pressure \( P = 101300 \, \text{Pa} \), and temperature \( T = 400 \, \text{K} \). Rearranging for the molar mass \( M \), we get \( M = \frac{mRT}{PV} \). Substituting \( R = 8.31 \, \text{J mol}^{-1} \text{K}^{-1} \), the expression simplifies to \( M = \frac{0.10 \times 8.31 \times 400}{101300 \times 85 \times 10^{-6}} \), matching option C.
X is an impure sample of a Group 2 metal carbonate, MCO₃. X contains 57% by mass of MCO₃. The impurities in X do not react with hydrochloric acid. 7.4 g of X is reacted with an excess of dilute hydrochloric acid. 0.050 mol of the Group 2 metal chloride is produced. What is the identity of the Group 2 metal?
▶️ Answer/Explanation
Ans: A
Given that 7.4 g of impure X contains 57% MCO₃, the mass of MCO₃ is \(7.4 \times 0.57 = 4.218 \text{ g}\). The reaction produces 0.050 mol of MCl₂, meaning 0.050 mol of MCO₃ reacted. The molar mass of MCO₃ is \(\frac{4.218 \text{ g}}{0.050 \text{ mol}} = 84.36 \text{ g/mol}\). Subtracting the mass of CO₃ (60 g/mol), the atomic mass of M is \(84.36 – 60 = 24.36 \text{ g/mol}\), which corresponds to Magnesium (Mg).
In this question it should be assumed that nitrogen behaves as an ideal gas under the conditions stated. Which volume is occupied by 1.00 g of nitrogen at 50.0°C and at a pressure of 120 kPa?
▶️ Answer/Explanation
Ans: B
Using the ideal gas equation \( pV = nRT \), we first calculate the number of moles \( n \) of nitrogen: \( n = \frac{1.00 \text{g}}{28.0 \text{g/mol}} = 0.0357 \text{mol} \). Convert the temperature to Kelvin: \( T = 50.0 + 273.15 = 323.15 \text{K} \). Rearranging the equation to solve for volume \( V \), we get \( V = \frac{nRT}{p} = \frac{0.0357 \times 8.314 \times 323.15}{120 \times 10^3} \approx 0.799 \text{dm}^3 \). Thus, the correct answer is B.
A 200 cm³ sample of water has an amount of oxygen gas dissolved in it.
This amount of oxygen gas has a volume of 6.00 cm³ when measured at 1.00 × \(10^5\) Pa and 35 °C.
What is the concentration of oxygen gas in the water? (You should assume that oxygen behaves as an ideal gas.)
▶️ Answer/Explanation
Ans: B
Use the ideal gas law \(PV = nRT\) to find moles of O₂. Convert 35°C to Kelvin (308 K). \(n = \frac{PV}{RT} = \frac{(1.00 \times 10^5)(6.00 \times 10^{-6})}{(8.31)(308)}\) (volume in m³). Calculate n ≈ \(2.34 \times 10^{-4}\) mol. Concentration = moles/volume = \((2.34 \times 10^{-4}) / 0.200\) = \(1.17 \times 10^{-3}\) mol dm⁻³.