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'?

1 Answer
Jun 16, 2018

Answer:

#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

#"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:

#color(blue)(bar(ul(|color(white)(a/a)pV = nRTcolor(white)(a/a)|)))" "#

We can rearrange this equation to get

#n = (pV)/(RT)#

NTP is 20 °C and 1 atm. Thus,

#p = "1.0 atm"#
#V = 1.69 × 10^"-4"color(white)(l)"mL" = 1.69 × 10^"-7"color(white)(l)"L"#
#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#
#T = "(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"#

Step 3. Calculate Avogadro's number

#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"#