Answer:
Van't Hoff factor for AlCl₃ = 3 (Approx)
Explanation:
Given:
Number of observed particular = 1.79 M
Number of theoretical particular = 0.56 M
Find:
Van't Hoff factor for AlCl₃
Computation:
Van't Hoff factor for AlCl₃ = Number of observed particular / Number of theoretical particular
Van't Hoff factor for AlCl₃ = 1.79 M / 0.56 M
Van't Hoff factor for AlCl₃ = 3.19
Van't Hoff factor for AlCl₃ = 3 (Approx)
To solve this, we can use two equations.
t1/2 = ln 2 / λ = 0.693 / λ
where, t1/2 is half-life and λ is the decay constant.
t1/2 = 10 min = 0.693 / λ
Hence, λ = 0.693 / 10 min - (1)
Nt = Nο e∧(-λt)
Nt = amount of atoms at t =t time
Nο= initial amount of atoms
t = time taken
by rearranging the equation,
Nt/Nο = e∧(-λt) - (2)
From (1) and (2),
Nt/Nο = e∧(-(0.693 / 10 min) x 20 min)
Nt/Nο = 0.2500
Percentage of remaining nuclei = (nuclei at t time / initial nuclei) x 100%
= (Nt/Nο ) x 100%
= 0.2500 x 100%
= 25.00%
Hence, Percentage of remaining nuclei is 25.00%
Yes, 50 pennies plus 50 pennies equals 100 pennies minus 50pennies equals 50 pennies.
