Answer:
35Cl = 75.9 %
37Cl = 24.1 %
Explanation:
Step 1: Data given
The relative atomic mass of Chlorine = 35.45 amu
Mass of the isotopes:
35Cl = 34.96885269 amu
37Cl = 36.96590258 amu
Step 2: Calculate percentage abundance
35.45 = x*34.96885269 + y*36.96590258
x+y = 1 x = 1-y
35.45 = (1-y)*34.96885269 + y*36.96590258
35.45 = 34.96885269 - 34.96885269y +36.96590258y
0.48114731 = 1,99704989y
y = 0.241 = 24.1 %
35Cl = 34.96885269 amu = 75.9 %
37Cl = 36.96590258 amu = 24.1 %
At STP one mole of any gas occupies 22.4 L
moles NO2 = 99.0/22.4 = 4.42
mass NO2 = 4.42 mol x 46.0 g/mol=203 g
Mass of Cl₂ : 164.01 g
<h3>Further explanation</h3>
A mole is a number of particles(atoms, molecules, ions) in a substance
This refers to the atomic total of the 12 gr C-12 which is equal to 6.02.10²³, so 1 mole = 6.02.10²³ particles
Can be formulated :
N = n x No
N = number of particles
n = mol
No = 6.02.10²³ = Avogadro's number
mol Cl₂ :

mass Cl₂(MW=71 g/mol) :

Molarity = moles of solution/ liter of solution
4.32x10^2/20
432/20=
21.6 M