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

- calcium has greater affinity to bond with electron receptors than hydrogen does and hence, it has the ability to replace hydrogen in most of its compounds.

- lithium has 2 valence electrons in the product because it can make only 1 bond with hydrogen. beryllium has 4 valence electrons in the product because it can only make 2 bonds. it has fewer electrons than neon and can react with electron donors. each hydrogen has 2 valence electrons, the same as helium.

- the covalent bond formed between carbon and hydrogen is polar because carbon is more electronegative than hydrogen and it attracts the bonded electron pair closer to itself, thus giving hydrogen a partial positive charge.

- a hydrogen bond is a relatively weak bond that hydrogen atoms make with the electronegative atoms nitrogen, oxygen or fluorine.

so the best choice is a.) 2 calcium (ca) atoms

i believe it is a.) released into

explanation: