Question

The solution having lowest vapour pressure amongst the following is.
1M Glucose
1m Sucrose
1M NaCl
1M K2SO4

Answer 1

Answer : The correct option is, 1 M $K_2SO_4$

Solution :

Formula used for lowering of vapor pressure :

$\frac{p^o-p_s}{p^o}=i\times x_2$

where,

$p^o$ = vapor pressure of the pure solvent

$p_s$ = vapor pressure of the solution

$x_2$ = mole fraction of the solute

i = Van't Hoff factor

Now we have to calculate the Van't Hoff factor for the given solutions.

(a) $C_6H_{12}O_6$, glucose is a non-electrolyte solute that means they retain their molecularity, and not undergo association or dissociation.

So, Van't Hoff factor = 1

(b) $C_{12}H_{22}O_{11}$, sucrose is also a non-electrolyte solute that means they retain their molecularity, and not undergo association or dissociation.

So, Van't Hoff factor = 1

(c) The dissociation of $NaCl$ will be,

$NaCl\rightarrow Na^++Cl^-$

So, Van't Hoff factor = Number of solute particles = 1 + 1 = 2

(d) The dissociation of $K_2SO_4$ will be,

$K_2SO_4\rightarrow 2K^++SO_4^{2-}$

So, Van't Hoff factor = Number of solute particles = 2 + 1 = 3

A the relative lowering of vapor pressure depends only on the Van't Hoff factor. That means the higher the Van't Hoff factor, lower will be the relative lowering of vapor pressure and vice-versa.

Hence, the correct option is, 1 M $K_2SO_4$