In sulphur estimation, $2.0 \times 10^{-3}$ mol of an organic compound $(X)$ (molar mass $76 \text{ g mol}^{-1}$) gave $0.4813 \text{ g}$ of barium sulphate (molar mass $233 \text{ g mol}^{-1}$). The percentage of sulphur in the compound $(X)$ is . . . . . . $\% \times 10^{-1}$ (Nearest integer).

In an organic compound,phosphorus is estimated as:

$0.2 \ g$ of an organic compound was subjected to estimation of nitrogen by Dumas method in which volume of $N_2$ evolved (at $STP$) was found to be $22.400 \ mL$. The percentage of nitrogen in the compound is $.......$. [nearest integer] (Given: Molar mass of $N_2$ is $28 \ g \ mol^{-1}$. Molar volume of $N_2$ at $STP$ : $22.4 \ L \ mol^{-1}$ )

The transformation occurring in Dumas's method is given below:
$C_{2}H_{7}N + (2x + \frac{y}{2})CuO \rightarrow xCO_{2} + \frac{y}{2}H_{2}O + \frac{z}{2}N_{2} + (2x + \frac{y}{2})Cu$
The value of $y$ is ...... .(Integer answer)

$0.3 \ g$ of an organic compound gave $50 \ mL$ of nitrogen collected at $300 \ K$ and $715 \ mm$ pressure in Dumas method. Calculate the percentage of nitrogen in the compound (aqueous tension of water at $300 \ K$ is $15 \ mm$).

In the sulphur estimation,$0.20 \ g$ of a pure organic compound gave $0.40 \ g$ of barium sulphate. The percentage of sulphur in the compound is $.......... \times 10^{-1} \%$. $(Molar \ mass : O=16, S=32, Ba=137 \ g \ mol^{-1})$