Question

In each pair, identify the solution that will have a higher boiling point. explain. 1. 1.50 moles of lioh (strong electrolyte) and 3.00 moles of koh (strong electrolyte) each in 1.0 kg of water 2. 0.40 mole of al(no3)3 (strong electrolyte) and 0.40 mole of cscl (strong electrolyte) each in 1.0 kg of water

Answer 1

Answer:- (1) 3.00 moles of KOH in 1.0 kg of water (2) 0.40 mole of $Al(NO_3)_3$ in 1.0 kg of water.

Explanations:- Elevation in boiling point is directly proportional to the molality of the solution.  The equation used for this is written as:

$\Delta T_b=imk_b$

where, $\Delta T_b$ is the elevation in boiling point, i is the Van't hoff factor, m is the molality of the solution and $k_b$ is the molal elevation constant for the solvent used.

Van't hoff factor is the theoretical number of ions an ionic compound give when it breaks. For example NaCl breaks to give $Na^+$ and $Cl^-$ . So, the value of i for NaCl is 2.

In the first pair we have LiOH and KOH. LiOH  gives lithium ion and hydroxide ion, so the value of i for this is 2. Similarly, KOH gives potassium ion and hydroxide ion and the value of i for this is also 2.

So, for this pair the deciding parameter is the molality. Molality of KOH is higher than LiOH and so the boiling point will be higher for KOH.

For second pair, the molality is same for both the solutions so the deciding parameter here is the value of van't hoff factor. CsCl breaks to give cesium ion and chloride ion and so the value of i for this is 2. $Al(NO_3)_3$ breaks to give one $Al^+^3$ ion and three $NO_3^-$ ions. So, the value of i for this is 4.

Since the value of i is higher for aluminium nitrate, it's boiling point will be higher.

Answer 2

The working equation for this is:

Tbp,soln - Tbp,water = i*Kb*m
where
Kb for water is 0.512 °C/molal
m is the molality (mol solute/kg solvent)
i is the van't hoff factor which represents the number of ions dissociated for strong electrolytes
Tbp,water is the boiling point of water which is 100°C

1. 1.50 moles of lioh (strong electrolyte) and 3.00 moles of koh (strong electrolyte) each in 1.0 kg of water

i = 2 for LiOH and 2 for KOH
Then,
Tbp,soln - 100 = (2+2)(0.512)((1.5+3)/1 kg)
Tbp,soln = 109.22°C

2. 0.40 mole of al(no3)3 (strong electrolyte) and 0.40 mole of cscl (strong electrolyte) each in 1.0 kg of water

i = 4 for al(no3)3 and 2 for cscl
Then,
Tbp,soln - 100 = (4+2)(0.512)((0.4+0.4)/1 kg)
Tbp,soln = 102.46°C

Thus, the first solution will have a higher boiling point.