JEE MAIN - Chemistry (2024 - 30th January Evening Shift - No. 8)
If a substance '$$A$$' dissolves in solution of a mixture of '$$B$$' and '$$C$$' with their respective number of moles as $$\mathrm{n}_{\mathrm{A}}, \mathrm{n}_{\mathrm{B}}$$ and $$\mathrm{n}_{\mathrm{C}_3}$$. Mole fraction of $$\mathrm{C}$$ in the solution is
$$\frac{n_C}{n_A \times n_B \times n_C}$$
$$\frac{n_B}{n_A+n_B}$$
$$\frac{n_C}{n_A+n_B+n_C}$$
$$\frac{n_C}{n_A-n_B-n_C}$$
Explanation
Mole fraction of $$\mathrm{C=\frac{n_C}{n_A+n_B+n_C}}$$
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