Answer:
Option B. 4.38 g
Step-by-step explanation:
The following data were obtained from the question:
Half life (t½) = 20 mins
Original amount (N₀) = 70 g
Time (t) = 80 mins
Amount remaining (N) =.?
Next, we shall determine the decay constant. This can be obtained as illustrated below:
Half life (t½) = 20 mins
Decay constant (K) =.?
Decay constant (K) = 0.693/ half life (t½)
K = 0.693/20
K = 0.03465 min¯¹
Finally, we shall determine the amount remaining after 80 mins as follow:
Original amount (N₀) = 70 g
Time (t) = 80 mins
Decay constant (K) = 0.03465 min¯¹
Amount remaining (N) =.?
Log (N₀/N) = kt/2.303
Log (70/N) = (0.03465 × 80)/2.303
Log (70/N) = 2.772/2.303
Log (70/N) = 1.2036
Take the anti log of 1.2036
70/N = antilog (1.2036)
70/N = 15.98
Cross multiply
70 = N × 15.98
Divide both side by 15.98
N = 70/15.98
N = 4.38 g
Therefore, the amount of the isotope remaining after 80 mins is 4.38 g