Question

which has the higher boiling point - 0.1M of NaCl or 0.1M glucose

Answer 1

BP is directly proportional to no. of free particles
In NaCl =2 (Na+,Cl-)
In glucose =1 (can't ionize)
so NaCl has higher BP

Answer 2

Answer : 0.1 M of NaCl has the higher boiling point.

Explanation :

Formula used for Elevation in boiling point :

$\Delta T_b=i\times k_b\times m$

where,

$\DeltaT_b$ = change in boiling point

$k_b$ = boiling point constant

m = molality

i = Van't Hoff factor

As per question, we conclude that the molality of the given solution are same. So, the boiling point depends only on the 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, an not undergo association or dissociation.

So, Van't Hoff factor = 1

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

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

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

The boiling point depends only on the Van't Hoff factor.That means higher the Van't Hoff factor, higher will be the boiling point and vice-versa.

Hence, 0.1 M of NaCl has the higher boiling point.