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# Which one of the following aqueous solutions has the lowest vapour pressure?

$\begin{array}{1 1}(a)\;0.1M \;glucose\\(b)\;0.1M \;Na_2SO_3\\(c)\;0.1M\;NaCl\\(d)\;0.1M\;sugar\end{array}$ Comment
A)
Toolbox:
• The greater the number of particles found in a solution the lower the vapor pressure.
• Colligative properties depend only the number of particles in solution, not their size, mass, charge, etc. You must determine the concentration of PARTICLES in order to determine the greatest deviation, and lowest vapor pressure of solvent.
• The van 't Hoff factor i (named after J. H. van 't Hoff) is a measure of the effect of a solute upon colligative properties such as osmotic pressure, relative lowering in vapor pressure, elevation of boiling point and freezing point depression. If a fraction $\alpha$ of the solute dissociates into n ions, then, $i=1+ \alpha (n-1)$
Since, glucose is a no-electrolyte. It doesn't dissociate in water. $C_6H_{12}O_6 (s) \longrightarrow C_6H_{12}O_6 (aq)$.
Thus, Van't Hoff factor of 0.1 M glucose is $i=1$
Thus, particle concentrationof glucose $= i \times Molarity = 1 \times 0.1 M = 0.1 M$
Next, $Na_2SO_3 \longrightarrow 2Na^+ + SO_3^{2-}$, Thus, $i=3$
$\therefore$Particle concentration of 0.1 M $Na_2SO_3$ is $3 \times 0.1M = 0.3M.$
$NaCl \longrightarrow Na^+ + Cl^-$, Thus, $i=2$
0.1 M NaCl has particle concentration of 0.2 M ( 2 $\times$ 0.1 M )
0.1 M sugar has particle concentration of 0.1 M.
Vapour pressure is inversely proportional to the 'i' value.
Since, 'i' value for $Na_2SO_3$ is high in comparision to others.$Na_2SO_3$ has the lowest vapour pressure.
Hence (b) is the correct answer.