Verified Solution • 2025–26 Session moderate

QJEEM23Chb703: JEE Main 2023 Chemistry - Class 12 JEEM23Chb703 Solution: Evaluate the following statements for their correctness. A. The …

Evaluate the following statements for their correctness.

A. The elevation in boiling point temperature of water will be same for 0.1M NaCl0.1 \mathrm{M} \, \mathrm{NaCl} and 0.1M0.1 \mathrm{M} urea.

B. Azeotropic mixtures boil without change in their composition.

C. Osmosis always takes place from hypertonic to hypotonic solution.

D. The density of 32% H2SO432 \% \, \mathrm{H}_{2} \mathrm{SO}_{4} solution having molarity 4.09 M4.09 ~\mathrm{M} is approximately 1.26 g mL−11.26 \mathrm{~g} \mathrm{~mL}^{-1}

E. A negatively charged sol is obtained when KI solution is added to silver nitrate solution.

Choose the correct answer from the options given below :

Multiple Choice Options

  • A A, B and D only
  • B A and C only
  • C B and D only
  • D B, D and E only

Step-by-Step Verified Solution

(A) Elevation in boiling point temperature of water will be higher for 0.1 M NaCl as compared to 0.1 M urea.

(B) Azeotropic mixtures boil without change in their composition

(C) Osmosis always takes place from hypotonic (low concentration of solute) solution to hypertonic (high concentration of solute) solution.

(E) When KI solution is added to AgNO3 solution, positively charged solution is formed due to adsorption of Ag+ ions from dispersion medium.

AgI/ Ag+ (positively charged)

(D) Let the mass of H2SO4 (32%) is 100.

∴ Weight of H2SO4=32\therefore \text { Weight of } \mathrm{H}_2 \mathrm{SO}_4=32

Moles of H2SO4=3298\mathrm{H}_2 \mathrm{SO}_4=\frac{32}{98}

Now, 4.09=3298×V⇒V=79ml4.09=\frac{32}{98 \times \mathrm{V}} \Rightarrow \mathrm{V}=79 \mathrm{ml}

Density =10079=1.265=\frac{100}{79}=1.265

Hence, correct answer is (C) B and D only.

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