Question

What is the value of the van't Hoff factor for KCl if a 1.00m aqueous solution shows a vapor pressure depression of 0.734 mmHg at 298 ∘C? (The vapor pressure of water at 298 K is 23.76 mmHg.)

Answer 1

Answer: The Van't Hoff factor for KCl is 1.74

Explanation:

We are given:

Molality of solution = 1 m

This means that 1 mole of a solute is present in 1 kg of solvent (water) or 1000 grams of water

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of water = 1000 g

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

$\text{Moles of water}=\frac{1000g}{18g/mol}=55.56mol$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

Moles of solute = 1 moles

Total moles = [1 + 55.56] = 56.56 moles

Putting values in above equation, we get:

$\chi_{(solute)}=\frac{1}{56.56}=0.0177$

The equation used to calculate relative lowering of vapor pressure follows:

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

where,

$\frac{p^o-p_s}{p^o}$ = relative lowering in vapor pressure = 0.734 mmHg

i = Van't Hoff factor = ?

$\chi_{solute}$ = mole fraction of solute = 0.0177

$p^o$ = vapor pressure of pure water = 23.76 torr

Putting values in above equation, we get:

$\frac{0.734}{23.76}=i\times 0.0177\\\\i=1.74$

Hence, the Van't Hoff factor for KCl is 1.74