2007 AJC Prelims Paper 1 FAQs

Explanation for Q1
Amt of Cl2 molecules in 1 dm3 of air at max toleration level
= (0.005 / 10^3) g / (2 x 35.5) g/mol

Amt of Cl2 molecules in 500 cm3 (which is 0.5 dm3) of air at max toleration level
= (0.005 / 10^3) g / (2 x 35.5) g/mol x 0.5

Amt of Cl atoms in 500 cm3 of air at max toleration level
= (0.005 / 10^3) g / (2 x 35.5) g/mol x 0.5 x 2 (because 1 Cl2 molecule has 2 Cl atoms)

No. of Cl atoms in 500 cm3 of air at max toleration level
= (0.005 / 1000) x (1/71) x 6 x 10^23

Explanation for Q3
For He, n(He) = (p1V1)/RT = (2 kPa x 1 dm3)/RT
For Ne, n(Ne) = (p2V2)/RT = (1 kPa x 2 dm3)/RT

Let new total pressure be p (in kPa).

p(1 dm3 + 2 dm3) = [n(He) + n(Ne)]RT
p(3 dm3) = [2 kPa dm3 + 2 kPa dm3]/RT x RT
p = 4/3 kPa