When 70.4 g of benzamide (C7H7NO) 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. Required:
Calculate the van't Hoff factor for ammonium chloride in X.

1.62

Explanation:

From the given information:

number of moles of benzamide  

= 0.58 mole

The molality =

= 0.6837

Using the formula:

where;

dT = freezing point = 27

l = Van't Hoff factor = 1

kf = freezing constant of the solvent

∴

2.7 °C = 1 × kf ×  0.6837 m

kf = 2.7 °C/ 0.6837m

kf = 3.949 °C/m

number of moles of NH4Cl =

= 1.316 mol

The molality =

= 1.5484

Thus;

the above kf value is used in determining the  Van't Hoff factor for  NH4Cl

i.e.

9.9 = l × 3.949 × 1.5484 m

l = 1.62

because u need two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei to form isotopes.

so since they all have the same shape? i mean maybe because m& m and reeses are both chocolates