Respuesta :
Answer : The molar mass of unknown compound is 152.38 g/mole
Explanation :
Depression in freezing point = [tex]1.9^oC[/tex]
Mass of unknown compound = 16.5 g
Mass of water = 106.0 g
Formula used :
[tex]\Delta T_f=i\times K_f\times m\\\\\Delta T_f=i\times K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of water in Kg}}[/tex]
where,
[tex]\Delta T_f[/tex] = depression in freezing point
i = Van't Hoff factor = 1 (for non-electrolyte)
[tex]K_f[/tex] = freezing point constant for water = [tex]1.86^oC/m[/tex]
m = molality
Now put all the given values in this formula, we get
[tex]1.9^oC=1\times (1.86^oC/m)\times \frac{16.5g\times 1000}{\text{Molar mass of unknown compound}\times 106.0g}[/tex]
[tex]\text{Molar mass of unknown compound}=152.38g/mole[/tex]
Therefore, the molar mass of unknown compound is 152.38 g/mole
Answer: The molar mass of the unknown compound is 152 g/mol.
Explanation:
Depression in freezing point is given by:
[tex]\Delta T_f=i\times K_f\times m[/tex]
[tex]\Delta T_f=T_f-^0T_f=(0-(-1.9)^0C=1.9^0C[/tex] = Depression in freezing point
i= vant hoff factor = 1 (for non electrolyte)
[tex]K_f[/tex] = freezing point constant = [tex]1.86^0C/m[/tex]
m= molality
[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]
Weight of solvent (water)= 106.0 g = 0.106 kg
Molar mass of unknown non electrolyte = M g/mol
Mass of unknown non electrolyte added = 16.5 g
[tex]1.9=1\times 1.86\times \frac{16.5g}{M g/mol\times 0.106kg}[/tex]
[tex]M=152g/mol[/tex]
Thus the molar mass of the unknown compound is 152 g/mol.
