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.

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.