The molar mass of the water insoluble product formed from the fusion of chromite ore $$(FeCr_{2}O_{4})$$ with $$Na_{2}CO_{3}$$ in presence of $$O_{2}$$ is_______$$gmol^{-1}$$.

The chromite ore is given as $$FeCr_{2}O_{4}$$. When fused with $$Na_{2}CO_{3}$$ in the presence of oxygen $$(O_{2})$$, the reaction occurs as follows:

The balanced chemical equation is:

$$4FeCr_{2}O_{4} + 8Na_{2}CO_{3} + 7O_{2} \rightarrow 8Na_{2}CrO_{4} + 2Fe_{2}O_{3} + 8CO_{2}$$

The products are sodium chromate $$(Na_{2}CrO_{4})$$, ferric oxide $$(Fe_{2}O_{3})$$, and carbon dioxide $$(CO_{2})$$.

Among these:

• Sodium chromate $$(Na_{2}CrO_{4})$$ is soluble in water.
• Carbon dioxide $$(CO_{2})$$ is a gas and soluble in water to some extent, but it is not a solid insoluble product.
• Ferric oxide $$(Fe_{2}O_{3})$$ is insoluble in water and remains as a solid.

Thus, the water-insoluble product is ferric oxide $$(Fe_{2}O_{3})$$.

To find the molar mass of $$Fe_{2}O_{3}$$:

Atomic mass of iron $$(Fe) = 56$$ g/mol.

Atomic mass of oxygen $$(O) = 16$$ g/mol.

Molar mass of $$Fe_{2}O_{3} = (2 \times 56) + (3 \times 16) = 112 + 48 = 160$$ g/mol.

Therefore, the molar mass of the water-insoluble product is 160 g mol⁻¹.