The equivalent weight of $H_3PO_4$ in the reaction is $Ca(OH)_2 + H_3PO_4 \to CaHPO_4 + 2H_2O$
$(Ca = 40, P = 31, O = 16)$

$8 \ g$ of a metal reacted with oxygen to form $9 \ g$ of its oxide. What is the weight (in $g$) of the metal required to react with $8 \ g$ of hydrogen? $(H=1.0)$

Isotope	Relative abundance $(\%)$	Atomic mass $(u)$
$^{12}C$	$98.8$	$12$
$^{13}C$	$1.18$	$13.1$
$^{14}C$	$0.02$	$14.1$

From the above data,what is the average molecular mass of $CH_4$ containing all isotopes of carbon,assuming hydrogen is only $^{1}_{1}H$? (Given that atomic mass of hydrogen $= 1.008 \ u$)

Assuming chlorine element has two isotopes as $Cl^{35}$ and $Cl^{37}$ with their percentage abundance as $25\%$ and $75\%$ respectively. Find the average atomic weight.

What is meant by $1 \ amu$ or $1 \ u$?

In chlorine gas,the ratio of $Cl^{35}$ and $Cl^{37}$ is