JEE Advanced 2022 Paper 1 · Chemistry · Chemical Kinetics · hard

A uranium-bearing mineral is analysed. The sample contains 1.19 g of ^238U, and it also traps 224 mL of helium (measured at 273 K and 1 atm; molar…

Numerical answer

A uranium-bearing mineral is analysed. The sample contains

1.19\ \mathrm{g}

^{238}\mathrm{U}

, and it also traps

224\ \mathrm{mL}

of helium (measured at

273\ \mathrm{K}

and

1\ \mathrm{atm}

; molar volume

22.4\ \mathrm{L\,mol^{-1}}

), all of which came from the complete decay of

^{238}\mathrm{U}

to stable

^{206}\mathrm{Pb}

through the radioactive series. Assume no helium has escaped and no lead was present initially. The half-life of

^{238}\mathrm{U}

4.5\times10^{9}

years. The age of the mineral, in units of

10^{9}

years, is (give the answer correct to two decimal places)

Show answer and solution

Answer: 1.45

Each atom of

^{238}\mathrm{U}

that reaches

^{206}\mathrm{Pb}

emits

\alpha

particles:

(238-206)/4=8

, with

6

\beta^-

particles to restore the charge (

92-16+6=82

). Helium moles

=0.224/22.4=0.0100

, so lead formed

=0.0100/8=1.25\times10^{-3}

mol. Uranium now present

=1.19/238=5.00\times10^{-3}

mol. Initial uranium

N_0=5.00\times10^{-3}+1.25\times10^{-3}=6.25\times10^{-3}

mol, so

N/N_0=0.80

. For first-order decay

t=\dfrac{t_{1/2}}{\ln 2}\ln\dfrac{N_0}{N}=4.5\times\dfrac{\ln1.25}{\ln2}=4.5\times0.3219=1.449

. The age is

1.45\times10^{9}

years. Treating each He as one-per-Pb (instead of eight) would give a wrong, much larger age. Answer: 1.45.

Original question written in the JEE Advanced 2022 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake