When 70.4 g of benzamide (c, h,no) are dissolved in 850. g of a certain mystery liquid x, the freezing point of the solution is 2.7 c lower than the freezing point of pure x. on the other hand, when 70.4 g of ammonium chloride (nh ci) are dissolved in the same mass of x, the freezing point of the solution is 9.9 °c lower than the freezing point of pure x. calculate the van't hoff factor for ammonium chloride in x. be sure your answer has a unit symbol, if necessary, and round your answer to 2 significant digits.

1.60 is the van't Hoff factor for ammonium chloride in X.

Explanation:

...(1)

where,

=Elevation in boiling point =

i = van't Hoff factor

= Freezing point constant

m = molality

1) When 70.4 g of benzamide  are dissolved in 850. g of a certain mystery liquid X.

Mass of benzamide = 70.4 g

Molar mass of benzamide = 121 g/mol

i = 1 (organic molecule)

Mass of liquid X = 850 g = 0.850 kg

= Freezing point constant of liquid X= ?

Putting all value in a (1):

2) When 70.4 g of ammonium chloride are dissolved in 850. g of a certain mystery liquid X.

Mass of ammonium chloride= 70.4 g

Molar mass of ammonium chloride = 53.5 g/mol

i = ?  (ionic molecule)

Mass of liquid X = 850 g = 0.850 kg

Putting all value in a (1):

i = 1.6011 ≈ 1.60

1.60 is the van't Hoff factor for ammonium chloride in X.