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 of Hg at $25^oC$. Further 18 g of water is then added to the solution, the new vapour pressure becomes 22.15 mm Hg at $25^oC$. Calculate : (a) Molar mass of solute (b) Vapour pressure of water at $25^oC$

  1. $11.39 \ mm$
  2. $22.8 \ mm$
  3. $23.78 \ mm$
  4. $45.6  \ mm$
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C Correct answer
AI explanation

Using Raoult's law for the two solutions, the equations 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 22.15 equals the vapor pressure of water multiplied by (108 divided by M) divided by (108 divided by M plus 30). Solving these two simultaneous equations for the unknown molar mass M and the vapor pressure of water gives M equal to 58.96 g/mol and the vapor pressure of water as 23.78 mm Hg.

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