¿a shaker of salt substitute contains 1.6 oz of k. what is the activity, in millicuries, of the potassium in the shaker? the activity is 7 microcuries (µci)

We know K-40 (potassium having atomic mass 40 g) is radioactive and its natural abundance is 0.012% 
So for 1 mol of potassium contains 0.00012 mol of K-40 
Now 1.6 oz of K = 45.36 g of K
average atomic weight of K = 39.1 g
so 45.36 g of K contains:
(45.36 / 39.1) * 0.00012 * 6.022 x 10²³ (atoms of K-40)
= 8.4 x 10¹⁹ atoms of K-40
We know, activity A is:
A = 0.693 / t1/2 N₀
[t1/2 : half life time and N₀ : initial number of atoms]
t1/2 of K-40 = 1.28 x 10⁹ years
    = 4.04 x 10¹⁶ seconds
So A = (0.693 / 4.04 x 10¹⁶) * (8.4 x 10¹⁹ ) = 1441 cps
A = 141 x (1/3.7 x 10¹⁰)   because 3.7 x 10¹⁰ cps = 1 Ci
A = 3.9 x 10⁻⁸ Ci  = 3.9 x 10⁻⁵ millicurie

Ithink d hope that