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

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The boiling point of water in a 0.1 molal silver nitrate solution (solution A ) is x ^ C . To this solution A , an equal volume of 0.1 molal aqueous barium chloride solution is added to make a new solution B. The difference in the boiling points of water in the two solutions A and B is y 10^ -2 ^ C . (Assume: Densities of the solutions A and B are the same as that of water and the soluble salts dissociate completely. Use: Molal elevation constant (Ebullioscopic Constant), K_ b =0.5 ~K ~kg ~mol ^ -1 ; Boiling point of pure water as 100^ C .) The value of  y is ___.

Quantrex Academy · Accepted Answer

Correct answer: A. A

After adding  BaCl 2  solution in  AgNO 3 solution  AgCl  will be precipitate. Since volume of both solutions are equal let's take V = 1 &#160; L  of each. so initial moles of  AgNO 3 = 0 .1 &#160; mole initial moles of  BaCl 2 = 0 .1 &#160; mole 2 &#160; AgNO 3 + BaCl 2 &#8594; 2 AgCl &#8595; + Ba NO 3 2 2 AgNO 3 + BaCl 2 &#8594; 2 AgCl &#8595; + Ba NO 3 2 Intial &#160; moles 0 .1 0 .1 Change 0 .1 &#8722; 0 .05 + 0 .1 + 0 .05 Final &#160; moles 0 0 .05 0 .05 Final &#160; concentration 0 .05 2 0 .05 2 New  &#916;T b = 1 &#160; K b &#160; m 3 &#215; 0 .05 2 + 3 &#215; 0 .05 2 &#215; 0 .5 = 0 .075 &#8728; C Difference in boiling point of both solution = &#916;T b 1 &#8722; &#916;T b 2 = 0 .1 &#8722; 0 .075 = 0 .025 = 2 .5 &#215; 10 &#8722; 2 K

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