Question

18g glucose is dissolved in $180g$ of water. The relative lowering of vapour pressure is ?
A) $1$
B) $1.8$
C) $0.01$
D) $0.001$

Answer 1

This lowering in vapour pressure is due to the fact that after the solute is added to the pure liquid (solvent), the liquid surface has molecules of both, the pure liquid and the solute. The number of solvent molecules that escape into the vapour phase gets reduced and as a result the pressure exerted by the vapour phase is also reduced. This is known as the relative lowering of vapour pressure.
${x_2} =$ mole fraction of the solute = $\dfrac{{{n_2}}}{{{n_1} + {n_2}}}$

Complete answer:
Molecular mass of water $= 18g$
Given mass= $180g$
Molecular mass of glucose $= 180g$
Given mass $= 18g$
To find: Relative lowering of vapour pressure
We first need to find the number of moles of glucose and water
Number of moles of glucose ( ${n_1}$ ) $= \dfrac{{18}}{{180}} = 0.10$
Number of moles of water ( ${n_2}$ ) $= \dfrac{{180}}{{18}} = 10$
Mole fraction of water( ${x_2}$ ) = $\dfrac{{{n_2}}}{{{n_1} + {n_2}}}$
Substituting the values in the above equation,
Mole fraction of water $= \dfrac{{10}}{{0.10 + 10}} = 0.990$
Relative Lowering of vapour pressure $= 1 - {x_2}$
Substituting the value of ${x_2}$ we get,
Relative Lowering of vapour pressure $= 1 - 0.990$
Relative Lowering of vapour pressure $= 0.01$
Hence, the correct option is C) $0.01$ .

Note:
Colligative properties are those of a solution that are dependent on the number of solute molecules present in the solution, regardless of their nature, in relation to the total number of molecules present in the solution. The relative lowering of vapour pressure, the depression of the freezing point, the elevation of the boiling point, and the osmotic pressure are all colligative properties that are beneficial for a solution.