40 mg diatomic volatile substance (X2) is converted to vapour that displaced 4.92 ml of air at 1atm and 300K.
FIND THE ATOMIC WEIGHT OF ELEMENT X.
Answer: 100
PLZ HELP GIVE ME THE METHOD
Answers
volume of vapor , V = 4.92ml = 4.92 × 10^-3 Litres
Temperature, T = 300K
pressure , P = 1atm
using ideal gas equation,
PV = nRT
or, 1 atm × 4.92 × 10^-3 Litres = n × 0.082 atm.litre/mol.K × 300K
or, 4.92 × 10^-3 = n × 0.082 × 300
or, n = 4.92 × 10^-3/(0.082 × 300)
= 0.2
hence, number of mole , n = 0.2
number of mole = given weight/molecular weight
or, 0.2 = 40mg/Molecular weight
or, molecular weight = 40/0.2 = 200mg
now atomic weight = molecular weight/2
= 200mg/2 = 100mg
Answer:
100mg
Explanation:
From the given data we have,
Mass of diatomic volatile substance X2 = 40mg
Volume of vapour displaced = 4.92ml
The temperature = 300K
Pressure per atom = 1atm
By implying the Ideal Gas Equation
Pressure X Volume = nRT
= 1 atm X 4.92 X 10^-3l
= n X 0.082 atm.L/mol.K X 300K
= 0.2 moles
To find the molecular weight
Molecular weight = X2/ total number of moles
= 40/0.2 = 200 mg
The atomic weight of the element
= 200mg/2 = 100mg