In the estimation of sulphur by Carius method,$0.468 \,g$ of an organic sulphur compound afforded $0.668 \,g$ of barium sulphate. Find out the percentage of sulphur in the given compound. (in $\%$)

During $S$ estimation,$160 \ mg$ of an organic compound gives $466 \ mg$ of barium sulphate. The percentage of Sulphur in the given compound is . . . . . . $\%.$ (Given molar mass in $g \ mol^{-1}$ of $Ba: 137, S: 32, O: 16$)

$A$ student has been given $0.314 \ g$ of an organic compound and asked to estimate Sulphur. During the experiment,the student has obtained $0.4813 \ g$ of barium sulphate. The percentage of sulphur present in the compound is . . . . . .
(Given Molar mass in $g \ mol^{-1}$: $S = 32$,$BaSO_4 = 233$) (in $\%$)

The reaction of $Q$ with $PhSNa$ yields an organic compound (major product) that gives a positive Carius test on treatment with $Na_2O_2$ followed by the addition of $BaCl_2$. The correct option$(s)$ for $Q$ is (are):

While estimating the nitrogen present in an organic compound by Kjeldahl's method,the ammonia evolved from $0.25 \ g$ of the compound neutralized $2.5 \ mL$ of $2 \ M \ H_2SO_4$. The percentage of nitrogen present in the organic compound is $......$

In $Duma's$ method for estimation of nitrogen,$0.25 \ g$ of an organic compound gave $40 \ mL$ of nitrogen collected at $300 \ K$ temperature and $725 \ mm$ pressure. If the aqueous tension at $300 \ K$ is $25 \ mm$,the percentage of nitrogen in the compound is: