Question

The boiling point of $Kr$ and $Rn$ are $-152^{\circ }C$ and $-62^{\circ }C$ respectively. Find the approximate boiling point of $Xe$

Answer 1

Answer: -107

According to the law of triad, the properties of middle element is average of rest two
$\therefore$ Boiling point of $Xe$ in a triad of $Kr,Xe,Rn$ =
(boiling point of $Kr$ + boiling point of Rn) / 2
$=(-152-62)/2=-107{}^{\circ }C$