Statement A: Elevation in boiling point for $0.1\,\mathrm{M}$ NaCl and $0.1\,\mathrm{M}$ urea

Elevation in boiling point is given by $\Delta T_b = i K_b m$ where $i$ is the van’t Hoff factor.

• For urea, $i=1$ (non-electrolyte).
• For NaCl, ideally $i\approx 2$ because it dissociates into $\mathrm{Na^+}$ and $\mathrm{Cl^-}$.

Therefore, for the same concentration, NaCl produces a greater elevation in boiling point than urea.

So, Statement A is false.

Statement B: Azeotropic mixtures boil without change in composition

By definition, an azeotrope is a mixture whose liquid and vapour compositions are the same at its boiling point. Hence it distills/boils without change in composition.

So, Statement B is true.

Statement C: Osmosis always takes place from hypertonic to hypotonic solution

Osmosis is the flow of solvent through a semipermeable membrane from lower solute concentration (hypotonic) to higher solute concentration (hypertonic).

The statement says the reverse.

So, Statement C is false.

Statement D: Density of $32\%\,\mathrm{H_2SO_4}$ solution with molarity $4.09\,\mathrm{M}$

Let us calculate density.

• $4.09\,\mathrm{M}$ means $4.09$ moles of $\mathrm{H_2SO_4}$ in $1\,\mathrm{L}$ solution.
• Molar mass of $\mathrm{H_2SO_4} = 98\,\mathrm{g\,mol^{-1}}$.

Mass of $\mathrm{H_2SO_4}$ in $1\,\mathrm{L}$ solution: $4.09 \times 98 = 400.82\,\mathrm{g}$

The solution is $32\%$ by mass, so $\frac{\text{mass of solute}}{\text{mass of solution}} = \frac{32}{100}$

Hence mass of solution is $\text{mass of solution} = \frac{400.82 \times 100}{32} = 1252.56\,\mathrm{g}$

Since this is for $1\,\mathrm{L} = 1000\,\mathrm{mL}$ solution, $\text{density} = \frac{1252.56}{1000} = 1.25256\,\mathrm{g\,mL^{-1}}$

This is approximately $1.25 \approx 1.26\,\mathrm{g\,mL^{-1}}$

So, Statement D is true.

Statement E: Negatively charged sol when KI is added to silver nitrate solution

For formation of AgI sol: $\mathrm{AgNO_3 + KI \rightarrow AgI(sol) + KNO_3}$

The charge on the sol depends on the ion in excess:

• If KI is added to $\mathrm{AgNO_3}$ solution, generally $\mathrm{AgNO_3}$ is in excess, so $\mathrm{Ag^+}$ gets adsorbed.
• Thus the sol becomes positively charged.

A negatively charged AgI sol is obtained when $\mathrm{AgNO_3}$ is added to KI solution, so that $\mathrm{I^-}$ is in excess.

Therefore, Statement E is false.

Final evaluation

• A: False
• B: True
• C: False
• D: True
• E: False

Therefore, the correct set is: $\boxed{\text{B and D only}}$

This corresponds to Option C.