# Radium emits (8.4*10¹⁰) ⋉ particle/gm/s and thus produces 'He' gas at the rate of 1.69*10⁻⁴ ml/gm/day at NTP. So what is the value of 'avogadro number'?

Jun 16, 2018

N_text(A) = 1.0 × 10^24

#### Explanation:

Assume that you have 1 g of radium.

Then, you have an emission rate of 8.4 × 10^10 α particles per second per day and helium formation at a rate of 1.69 × 10^"-4" mL/day.

Step 1. Calculate the number of α particles produced in a day

$\text{No. of particles}$

= 1 color(red)(cancel(color(black)("day"))) × (24 color(red)(cancel(color(black)("h"))))/(1 color(red)(cancel(color(black)("day")))) × (60 color(red)(cancel(color(black)("min"))))/(1 color(red)(cancel(color(black)("h")))) × (60 color(red)(cancel(color(black)("s"))))/(1 color(red)(cancel(color(black)("min")))) × (8.4 × 10^10 color(white)(l)"α particles")/(1 color(red)(cancel(color(black)("s"))))

= 7.26 × 10^15color(white)(l)"α particles"

Step 2. Calculate the moles of helium formed in a day

We can use the Ideal Gas Law to solve this problem:

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} p V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

We can rearrange this equation to get

$n = \frac{p V}{R T}$

NTP is 20 °C and 1 atm. Thus,

$p = \text{1.0 atm}$
V = 1.69 × 10^"-4"color(white)(l)"mL" = 1.69 × 10^"-7"color(white)(l)"L"
$R = \text{0.082 06 L·atm·K"^"-1""mol"^"-1}$
$T = \text{(20 + 273.15) K" = "293.15 K}$

n = (1.0 color(red)(cancel(color(black)("atm"))) × 1.69 × 10^"-7" color(red)(cancel(color(black)("L"))))/("0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 293.15 color(red)(cancel(color(black)("K")))) = 7.03 × 10^"-9"color(white)(l)"mol"

N_text(A) = (7.26 × 10^15 color(white)(l)"particles")/(7.03 × 10^"-9" color(white)(l)"mol") = 1.0 × 10^24 color(white)(l)color(white)(l)"particles/mol"