To determine the correctness of the statements, we'll analyze each one individually by applying principles of solubility, common ion effect, and chemical equilibria.

Statement I:

On passing $\mathrm{HCl}_{(g)}$ through a saturated solution of $\mathrm{BaCl}_2$ at room temperature, white turbidity appears.

Analysis:

Dissolution of HCl Gas:

When $\mathrm{HCl}_{(g)}$ is bubbled through water, it dissolves and dissociates completely:

$ \mathrm{HCl}_{(aq)} \rightarrow \mathrm{H}^+_{(aq)} + \mathrm{Cl}^-_{(aq)} $

This increases the concentration of $\mathrm{Cl}^-$ ions in the solution.

Effect on BaCl₂ Solubility:

The solubility equilibrium of $\mathrm{BaCl}_2$ in water is:

$ \mathrm{BaCl}_2 \leftrightarrow \mathrm{Ba}^{2+}_{(aq)} + 2\mathrm{Cl}^-_{(aq)} $

Adding more $\mathrm{Cl}^-$ shifts the equilibrium to the left (Le Chatelier's Principle), causing $\mathrm{BaCl}_2$ to precipitate.

The precipitation of $\mathrm{BaCl}_2$ manifests as a white turbidity.

Conclusion:

Statement I is correct.

Statement II:

When $\mathrm{HCl}$ gas is passed through a saturated solution of $\mathrm{NaCl}$, sodium chloride is precipitated due to common ion effect.

Analysis:

Dissolution of HCl Gas:

Similar to before, $\mathrm{HCl}$ increases $\mathrm{Cl}^-$ ion concentration.

Effect on NaCl Solubility:

The solubility equilibrium of $\mathrm{NaCl}$ is:

$ \mathrm{NaCl} \leftrightarrow \mathrm{Na}^+_{(aq)} + \mathrm{Cl}^-_{(aq)} $

However, $\mathrm{NaCl}$ is highly soluble in water, and its solubility is not significantly affected by the common ion effect from $\mathrm{Cl}^-$.

The solubility product ($K_{sp}$) of $\mathrm{NaCl}$ is large, and the addition of $\mathrm{Cl}^-$ ions does not cause $\mathrm{NaCl}$ to precipitate under normal conditions.

No precipitation occurs; the solution remains clear.

Conclusion:

Statement II is incorrect.

Final Answer:

Statement I is correct but Statement II is incorrect.