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

your answer would be “dry”

the answer is 1260

you spend 7 hours in school for 1 school day with 5 school days in 1 school week with 36 school weeks in a school year.

7*5*36= 1260 it's all about multiplying