Question

In a sample of radioactive element, radium disintegrates at an average rate of $2.24\times {10}^{13}\alpha$-particles per minute. Each $\alpha$-particle takes up 2 electrons from the air and becomes a neutral helium atom. After 420 days, the He gas collected was 0.5 mL measured at ${27}^{\circ }C$ and 750 mm of mercury pressures. From the above data, Avogadro's number calculated is :

• A. $6.775\times {10}^{23}$ particles per mol
• B. $6.55\times {10}^{23}$particles per mol
• C. $6.470\times {10}^{23}$ particles per mol
• D. none of these

Answer 1

The correct option is A $6.775\times {10}^{23}$ particles per mol

Given that, No. of $\alpha$ particles (or He) formed $=2.24\times {10}^{13}minut{e}^{-1}$

No. of $\alpha$-particles formed in 420 day$=2.24\times {10}^{13}\times 420\times 24\times 60=1.355\times {10}^{19}$ ......(1)

Also, given at ${27}^{\circ }C$ and 750 mm of P, the volume of $He$ obtained is $0.5$ mL.
Using, the ideal gas equation $PV=nRT$
$\frac{750}{760}\times \frac{0.5}{1000}=n\times 0.0821\times 300$
$\therefore n=2.0\times {10}^{-5}mol$$=2.0\times {10}^{-5}\times N$ molecule ......(2) (here N is Avogadro No.)
Note: When the number of moles are multiplied by Avogadro number. the number of molecules is obtained.
Comparing the number of molecules of He formed in equations (1) and (2)
$1.355\times {10}^{19}=2.0\times {10}^{-5}\times N$
The Avogadro's number, $N=6.775\times {10}^{23}$ particles per mol