In sulphur estimation, 2.0 × 10^ -3 mol of an organic compound (X)...

Mass of the organic compound (X) can be calculated as:

$\text{Mass of X} = \text{Moles} \times \text{Molar mass}$

$\text{Mass of X} = 2.0 \times 10^{-3} \text{ mol} \times 76 \text{ g mol}^{-1} = 0.152 \text{ g}$

In the Carius method for sulphur estimation, the mass of sulphur in the given mass of $\text{BaSO}_4$ is:

$\text{Mass of S} = \dfrac{32}{233} \times \text{Mass of BaSO}_4$

$\text{Mass of S} = \dfrac{32}{233} \times 0.4813 \text{ g} \approx 0.0661 \text{ g}$

The percentage of sulphur in the organic compound is given by:

$\text{\% of S} = \dfrac{\text{Mass of S}}{\text{Mass of X}} \times 100$

$\text{\% of S} = \dfrac{0.0661}{0.152} \times 100 \approx 43.487 \text{ \%}$

To express this in the requested format:

$43.487 \text{ \%} = 434.87 \times 10^{-1} \text{ \%}$

Rounding to the nearest integer, we get $435$.

Answer: $435$