Skip to content
dronacharya.ai
← JEE Advanced 2021 · Paper 1
ChemistryNumerical+2Question 29 of 57

JEE Advanced 2021 · Paper 1, Chemistry question 10

Passage

Question Stem for Question Nos. 9 and 10

The boiling point of water in a 0.1 molal silver nitrate solution (solution A\mathbf A) is x\mathbf x °C. To this solution A\mathbf A, an equal volume of 0.1 molal aqueous barium chloride solution is added to make a new solution B\mathbf B. The difference in the boiling points of water in the two solutions A\mathbf A and B\mathbf B is y×10−2\mathbf y \times 10^{-2} °C.

(Assume: Densities of the solutions A\mathbf A and B\mathbf B are the same as that of water and the soluble salts dissociate completely.

Use: Molal elevation constant (Ebullioscopic Constant), Kb=0.5K_b = 0.5 K kg mol−1^{-1}; Boiling point of pure water as 100 °C.)

The value of ∣y∣|\mathbf y| is ____.

Answer and worked solution

Official answer: 2.5

After mixing equal volumes (densities as water, so molalities halve):

  • From A: AgX+\ce{Ag+} 0.05 and NOX3X−\ce{NO3-} 0.05.
  • From the BaClX2\ce{BaCl2} solution: BaX2+\ce{Ba^{2+}} 0.05 and ClX−\ce{Cl-} 0.10.

Precipitation. AgCl precipitates and removes 0.05 of AgX+\ce{Ag+} and 0.05 of ClX−\ce{Cl-}. Left in solution: NOX3X−\ce{NO3-} 0.05, BaX2+\ce{Ba^{2+}} 0.05 and ClX−\ce{Cl-} 0.05, a total of 0.15 mol kg−1^{-1} of ions.

Boiling points.

  • B: ΔTb=0.5×0.15=0.075\Delta T_b=0.5\times0.15=0.075 K.
  • A: ΔTb=0.1\Delta T_b=0.1 K.

The difference is 0.0250.025 °C =2.5×10−2=2.5\times10^{-2} °C, so ∣y∣=2.5|y|=2.5.

Topics:Colligative propertiesvan't Hoff factorSolubility product
← Previous questionNext question →