A gas evolved during the fermentation of sugar was collected at 22.5°C and 702 mmHg. After purification its volume was found to be 25.0 L. How many moles of gas were collected? (R = 0.08206 L • atm/K • mol, 1 atm = 760 mmHg)
A) 0.952 mol
B) 1.05 mol
C) 12.5 mol
D) 22.4 mol
E) 724 mol

Question

Quizplus · Accepted Answer

To find the moles of gas collected, we can use the ideal gas law equation: PV = nRT. First, we need to convert the given pressure from mmHg to atm by dividing by 760 mmHg/atm: 702 mmHg / 760 mmHg/atm = 0.924 atm. The temperature must be in Kelvin, so we convert 22.5°C to Kelvin by adding 273.15: 22.5°C + 273.15 = 295.65 K. Now, we can solve for n (moles of gas) using the values: n = PV / RT = (0.924 atm * 25.0 L) / (0.08206 L•atm/K•mol * 295.65 K) ≈ 0.952 mol.

A Gas Evolved During the Fermentation of Sugar Was Collected