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.)

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:

Given mass of water = 1000 g

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

Mole fraction of a substance is given by:

Moles of solute = 1 moles

Total moles = [1 + 55.56] = 56.56 moles

Putting values in above equation, we get:

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

where,

= relative lowering in vapor pressure = 0.734 mmHg

i = Van't Hoff factor = ?

= mole fraction of solute = 0.0177

= vapor pressure of pure water = 23.76 torr

Putting values in above equation, we get:

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

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

answercontent filtering;