When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezing point of pure . On the other hand, when of ammonium chloride are dissolved in the same mass of , the freezing point of the solution is lower than the freezing point of pure . Calculate the van't Hoff factor for ammonium chloride in . Be sure your answer has a unit symbol, if necessary, and is rounded to the correct number of significant digits.

The figures and parameters to answer this question are not clearly stated, so I will answer it in the closest best way I can.

Explanation:

Molality of alanine = mole / weight of solvent( kg)

= (170×1000)/(89×600)

= 3.18 molal

We know ∆ Tf = Kf × m

Kf = ∆ Tf / molality

= 7.9/3.18 = 2.48 °c.kg.mol-1

Now using the value of cryoscopic constant we calculate can't Hoff factor for Ammonium chloride.

i = ∆ Tf /( Kf × molality of NH4Cl)

= (24.7 × 53.5 × 600) /(2.48 × 170 × 1000)

= 1.8

the formula for the conversion of temperature from degree celsius to kelvin is:

the sample of water at is converted to kelvin temperature by using the above formula as:

on comparing the two samples of water at or we can say that at and at respectively. the difference between the temperatures of two samples of water is:

hence, one can say that the sample of water at is greater than the sample of water at by .