Question

The hydride of an unstable nuclide of a Group IIA metal, MH2(s), decays by a-emission. A 0.025-mol sample of the hydride is placed in an evacuated 1.1 −L container at 298 K. After 64 minutes, the pressure in the container is 0.49 atm.

Find the half-life of the nuclide.

Express your answer to two significant figures and include the appropriate units. t 1/2=___

The answer is not 21 minutes. Already tried it.

Answer 1

Given,

Initial moles of nuclide(N_o) = 0.025 mol

Also given,

Volume of container = 1.1 L

Temperature(T) = 298 K

Pressure(P) = 0.49 atm

Thus, From the given data, Calculating the number of moles of nuclide remained after 64 minutes in the evacuated container,

We know,

PV = nRT

Rearranging the formula,

n =PV/RT

n =( 0.49 atm x 1.1 L / 0.08206 L. atm/ mol K x 298 K)

n = 0.02204 mol

Thus, N_t = 0.02204 mol

Now, Calculating the decay constant(k),

ln [ N_o /N_t] = k x t

ln [ 0.025 / 0.022] = k x 64 min.

k = 0.001968 min^-1

Now, We know,

t_1/2 = 0.693 / k

t_1/2 = 0.693 / 0.001968 min^-1

t_1/2 = 352 min.

Thus, the half-life period for the nuclide is 3.5 x 10² minutes Or 350 minutes.