Question

40 mg of gaseous substance (X) occupies 4.8 mL of volume at 1 atm and 27°C. Atomic mass of element
X is : (R: 0.08 atm L/mole k​

Answer 1

volume of gaseous substance, V = 4.8mL ,

pressure of substance, P = 1atm

temperature , T = 27°C = 300K

weight of substance, w = 40mg = 0.04g

using gaseous formula, PV = nRT

or, 1atm × 4.8ml = n × 0.08atm.L/mol.K × 300K

or, 1atm × 4.8 × 10^-3 L = n × 0.08atm.L/mol.K × 300K

or, 4.8 × 10^-3/(0.08 × 300) = n

or, n = 2 × 10^-4 mol.

we know, mole of atom = given weight/atomic weight

so, n = 2 × 10^-4 = 0.04/atomic weight

or, atomic weight = 0.04 /2 × 10^-4

= 200 g/mol

hence, atomic weight of element is 200g/mol.