Question

A gas (Molar mass = 280 g mol^–1) was burnt in excess O₂ in a constant volume calorimeter and during combustion the temperature of calorimeter increased from 298.0 K to 298.45 K. If the heat capacity of calorimeter is 2.5 kJ K^–1 and enthalpy of combustion of gas is 9 kJ mol^–1 then amount of gas burnt is __________g. (Nearest Integer).

Answer 1

Correct Answer: 35

The problem requires calculating the amount of gas burnt in a calorimeter setup. Let's break down the information given and solve systematically:

1. Initial Data:

• Molar mass of the gas = 280 g/mol
• Temperature change (ΔT) = 298.45 K - 298.0 K = 0.45 K
• Heat capacity of calorimeter (C) = 2.5 kJ/K
• Enthalpy of combustion (ΔH_c) = 9 kJ/mol

2. Calculate the heat absorbed by the calorimeter:

• Using the formula for heat absorbed: q = CΔT = 2.5 kJ/K × 0.45 K = 1.125 kJ

3. Relate the heat absorbed to the amount of gas burnt:

• The calorimeter absorbs heat equivalent to the enthalpy of combustion of the gas.
• 1.125 kJ = (n moles) × (ΔH_c)
• n = 1.125 kJ / 9 kJ/mol = 0.125 mol

4. Calculate the mass of gas burnt:

• Mass = n × Molar mass = 0.125 mol × 280 g/mol = 35 g

5. Validation: The calculated mass (35 g) is within the provided range of (35,35).

Conclusion: The amount of gas burnt is 35 g.