In Dumas method,$0.3 \ g$ of an organic compound gave $45 \ mL$ of nitrogen at $STP$. The percentage of nitrogen is

Calculate the percentage of nitrogen in $0.25 \ g$ of an organic compound if $38 \ cm^3$ of $N_2$ gas is evolved at $300 \ K$ and $96 \ kPa$ pressure. (Given: $1 \ atm = 101.3 \ kPa$)

What is the reason for passing dry air through the combustion tube during the estimation of $C$ and $H$?

In Kjeldahl's method for the estimation of nitrogen,the formula used is

Consider the following reaction. On estimation of bromine in $1.00 \ g$ of $R$ using Carius method,the amount of $AgBr$ formed (in $g$) is . . . .
$\overline{[Given :}$ Atomic mass of $H = 1, C = 12, O = 16, P = 31, Br = 80, Ag = 108]$

$0.12 \ g$ of an organic compound '$X$' containing phosphorus on reacting with magnesia mixture gave $0.22 \ g$ of magnesium pyrophosphate $(Mg_2P_2O_7)$. The percentage of phosphorus in the compound '$X$' is (in $\%$)