Answer :
Answer: The molar mass of the unknown solute is 35.8 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^0-T_f=(0-(-2.60))^0C=2.60^0C[/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)=[tex]density\times volume =1.00g/ml\times 100ml=100g=0.1kg[/tex] (1kg=1000g)
Molar mass of unknown non electrolyte = M g/mol
Mass of unknown non electrolyte added = 5.00 g
[tex]2.60=1\times 1.86\times \frac{5.00g}{M g/mol\times 0.100kg}[/tex]
[tex]M=35.8g/mol[/tex]
The molar mass of the unknown solute is 35.8 g/mol