Which one of the following aqueous solutions will exhibit highest boiling point?

0.01 M N a_{2} S O_{4}

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0.01 M K N O_{3}

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0.015 M u r e a

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0.015 M g l u c o s e

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Solution

The correct option is A $0.01 M N a_{2} S O_{4}$
Formula used for Elevation in boiling point :

$Δ T_{b} = i \times k_{b} \times m$

where,
$Δ T_{b} =$ change in boiling point
$k_{b} =$ boiling point constant
m = molality
i = Van't Hoff factor

According to the formula, we conclude that to 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) The dissociation of $N a_{2} S O_{4}$ will be,

$N a_{2} S O_{4} \to 2 N a^{+} + {S O_{4}}^{2 -}$

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

(b) $C_{6} H_{12} O_{6}$ (glucose) is a non-electrolyte solute that means they retain their molecularity, an not undergo association or dissociation.

So, Van't Hoff factor = 1

(C) The dissociation of $C H_{4} N_{2} O$ (urea) is a non-electrolyte solute that means they retain their molecularity, an not undergo association or dissociation.

So, Van't Hoff factor = 1

(d) The dissociation of $K N O_{3}$ will be,

$K N O_{3} \to K^{+} + {N O_{3}}^{-}$

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

So, Van't Hoff factor = 2

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

Hence, the correct option is $A$

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