Respuesta :
Answer:
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 [tex]\Delta T_f= 8.2^oC[/tex]
Molality of the solution = m
The van't Hoff factor for glycine (non ionic) in liquid X= i =1
[tex]m=\frac{282 g}{75.07 g/mol\times 0.950 kg}=3.9542 mol/kg[/tex]
[tex]\Delta T_f=i\times K_f\times m[/tex]
[tex]8.2^oC=1\times K_f\times 3.9542 mol/kg[/tex]
The value of molal depression constant for liquid X:
[tex]K_f=2.0737 ^oC kg/mol[/tex]..(1)
Now 282. g of sodium chloride are dissolved in the same mass of X.
Depression in freezing point of the NaCl solution [tex]\Delta T_f'= 20.0^oC[/tex]
Molality of the NaCl solution = m'
The van't Hoff factor for NaCl(ionic) in liquid X= i'
[tex]m'=\frac{282 g}{58.5 g/mol\times 0.950 kg}=5.0742 mol/kg[/tex]
[tex]\Delta T_f'=i'\times K_f\times m'[/tex]
[tex]20.0^oC=i'\times 2.0737 ^oC kg/mol\times 5.0742 mol/kg[/tex]
i = 1.9007 ≈ 1.9
The van't Hoff factor for sodium chloride in X is 1.9 .
