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

To determine the correct answer for the metals $x$, $y$, and $z$ described in the question, let's analyze each metal chloride based on their solubility properties: 

1. $x$ is water soluble at room temperature:
   • Among the options provided, $\mathrm{CuCl_2}$ and $\mathrm{AlCl_3}$ are known to be soluble in water at room temperature.
2. $y$ is sparingly soluble in water at room temperature:
   • $\mathrm{AgCl}$ and $\mathrm{PbCl_2}$ are examples of chlorides that are sparingly soluble in water at room temperature.
   • Of the provided chlorides, $\mathrm{AgCl}$ has the least solubility, making it the likely candidate for $y$.
3. $z$ is soluble in hot water:
   • $\mathrm{PbCl_2}$ is more soluble in hot water than at room temperature.
   • This is a characteristic property of lead(II) chloride, which fits $z$.

Based on the analysis of solubility:

• $x$ should be $\mathrm{CuCl_2}$ because it is water-soluble at room temperature.
• $y$ is likely $\mathrm{AgCl}$ as it is sparingly soluble in water at room temperature.
• $z$ is $\mathrm{PbCl_2}$, soluble in hot water.

Therefore, the correct answer is: $\mathrm{CuCl_2, AgCl\ and\ PbCl_2}$.