All of the carbon dioxide in a soft drink with an initial mass of 381.04 g was carefully extracted and collected as a gas - HSC - SSCE Chemistry - Question 19 - 2011 - Paper 1

Using the molar mass of carbon dioxide (CO₂), which is approximately 44.01 g/mol:

Moles of CO₂ = ( \frac{0.63 \text{ g}}{44.01 \text{ g/mol}} \approx 0.0143 \text{ mol} )

The ideal gas law is given by the equation: ( PV = nRT ), where:

• ( P ) = pressure in kPa = 100 kPa
• ( V ) = volume in L
• ( n ) = moles of gas = 0.0143 mol
• ( R ) = ideal gas constant = 8.314 J/(mol·K) = 8.314 L·kPa/(mol·K)
• ( T ) = temperature in Kelvin = 25°C = 298 K

Rearranging the equation, we find: [ V = \frac{nRT}{P} = \frac{0.0143 \text{ mol} \times 8.314 \text{ L·kPa/(mol·K)} \times 298 \text{ K}}{100 \text{ kPa}} \approx 0.35 \text{ L} ]

The volume that the carbon dioxide would occupy at 100 kPa and 25°C is approximately 0.35 L. Therefore, the correct answer is (B) 0.35 L.