This problem has been solved! see the answerwhen 282. g of glycine (c2h5no2) are dissolved in 950. g of a certain mystery liquid x, the freezing point of the solution is 8.2c lower than the freezing point of pure x. on the other hand, when 282. g of sodium chloride are dissolved in the same mass of x, the freezing point of the solution is 20.0c lower than the freezing point of pure x. calculate the van't hoff factor for sodium chloride in x.

The van't Hoff factor for sodium chloride in X is 1.9 .

Explanation:

Mass of liquid x = 950 g = 0.950 kg

When 282 g of glycine are dissolved in 950 g of a certain mystery liquid X.

Depression in freezing point of the solution

Molality of the solution = m

The van't Hoff factor for glycine (non ionic) in liquid X= i =1

The value of molal depression constant for liquid X:

..(1)

Now 282. g of sodium chloride are dissolved in the same mass of X.

Depression in freezing point of the NaCl solution

Molality of the NaCl solution = m'

The van't Hoff factor for NaCl(ionic) in liquid X= i'

i = 1.9007 ≈ 1.9

The van't Hoff factor for sodium chloride in X is 1.9 .

answer carbon and three oxygen atoms;