Question

What is the mass of an original 5.60-gram sample of iron-53 that remains unchanged after 25.53 minutes?(1) 0.35g

(2) 1.40 g(3) 0.70 g
(4) 2.80 g

Answer 1

After 25.53 minutes there will remain 0.70 grams of the iron-53 sample. Option 3 is correct.

Step-by-step explanation:

Step 1: Data given

Original mass of the original iron sample = 5.60 grams

Half life of iron-53 = 8.5167 minutes

Step 2: Calculate the rate constant

k = (ln 2)/(t1/2)

⇒ with k = the rate constant

⇒ with t1/2 = the half-life time = 8.5167 minutes

k = 0.08139 / min

Step 3: Calculate the mass after 25.53 minutes

At = A0*e^(-kt)

⇒ with At = the amount of iron sample after 25.53 minutes

⇒ A0 = the initial amount of iron sample =5.50 grams

⇒ with k = 0.08139 / min

⇒ with t = the time = 25.53 minutes

At = 5.50*e^(-0.08139*25.53)

At = 0.70 grams

After 25.53 minutes there will remain 0.70 grams of the iron-53 sample

Answer 2

(3) 0.70 g

Step-by-step explanation:

Half life of iron-53 = 8.5167 minutes

`t_(1/2)=\frac {ln\ 2}{k}`

Where, k is rate constant

So,

`k=\frac {ln\ 2}{t_(1/2)}`

`k=(ln\ 2)/(8.5167)\ min^(-1)`

The rate constant, k = 0.08138 min⁻¹

Time = 25.53 minutes

Using integrated rate law for first order kinetics as:

`[A_t]=[A_0]e^(-kt)`

Where,

`[A_t]` is the concentration at time t

`[A_0]` is the initial concentration = 5.60 g

So,

`([A_t])/(5.60)=e^(-0.08138* 25.53)`

`[A_t]=(5.6)/(e^(2.0776314))`

`[A_t]=0.70\ g`

Answer- (3) 0.70 g