Question

Consider three metal chlorides $x$, $y$ and $z$, where $x$ is water soluble at room temperature, $y$ is sparingly soluble in water at room temperature and $z$ is soluble in hot water. $x$, $y$ and $z$ are respectively ___.

• A. $\mathrm{AlCl_3,\ PbCl_2\ and\ BaCl_2}$
• B. $\mathrm{AgCl,\ Hg_2Cl_2\ and\ PbCl_2}$
• C. $\mathrm{CuCl_2,\ AgCl\ and\ PbCl_2}$
• D. $\mathrm{MgCl_2,\ AgCl\ and\ AlCl_3}$

Hint:

Remember solubility trends of common chlorides: AgCl is sparingly soluble, PbCl$_2$ dissolves in hot water.

Answer 1

The Correct Option is C

Step 1: Identifying chloride soluble at room temperature.
Copper(II) chloride $(\mathrm{CuCl_2})$ is highly soluble in water at room temperature due to its ionic nature and hydration energy.
Step 2: Identifying sparingly soluble chloride.
Silver chloride $(\mathrm{AgCl})$ is sparingly soluble in water at room temperature because of its very low solubility product ($K_{sp}$).
Step 3: Identifying chloride soluble in hot water.
Lead(II) chloride $(\mathrm{PbCl_2})$ is sparingly soluble in cold water but becomes soluble in hot water due to increased kinetic energy of ions.
Step 4: Final conclusion.
Thus, $x = \mathrm{CuCl_2}$, $y = \mathrm{AgCl}$ and $z = \mathrm{PbCl_2}$, which corresponds to option (3).