Multiple choice

A solution containing $30:g$ of a non-volatile solute in exactly $90:g$ water has a vapour pressure of $21.85: mm: Hg$ at $25^{\circ}C$. Further 18 gm of water is then added to the solution. The resulting solution has a vapour pressure of 22.18 mmHg at $25^{\circ}C$. Calculate (a) molar mass of the solute, and (b) vapour pressure of water at $25^{\circ}C$.

  1. $\;(a)\:30.6\:g\:mol^{-1},\;(b)\:15.53\:mm\:Hg$
  2. $\;(a)\:61.21\:g\:mol^{-1},\;(b)\:23.99\:mm\:Hg$
  3. $\;(a)\:65.45\:g\:mol^{-1},\;(b)\:28.87\:mm\:Hg$
  4. $\;(a)\:84.87\:g\:mol^{-1},\;(b)\:39.98\:mm\:Hg$
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B Correct answer
Explanation

Raoult's Law: (P0 - P) / P0 = n2 / (n1 + n2). Let M be molar mass of solute. (P0 - 21.85) / P0 = (30/M) / (90/18 + 30/M) = (30/M) / (5 + 30/M) = 30 / (5M + 30). After adding 18g water: (P0 - 22.18) / P0 = (30/M) / (108/18 + 30/M) = (30/M) / (6 + 30/M) = 30 / (6M + 30). Solving the system for M and P0 yields M = 61.21 and P0 = 23.99.

AI explanation

Using Raoult's law, the two equations for the solutions are 21.85 equals the vapor pressure of pure water multiplied by (90 divided by M) divided by (90 divided by M plus 30), and 22.18 equals the vapor pressure of water multiplied by (108 divided by M) divided by (108 divided by M plus 30). Solving these simultaneous equations yields a molar mass of the solute equal to 61.21 g/mol and the vapor pressure of water at 25 degrees Celsius as 23.99 mm Hg.

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