Answer: 1. The molecular weight of the compound is 222.8 g/mol
2. The probable molecular formula of the solute is [tex]InCl_3[/tex]
Explanation:
Elevation in boiling point is given by:
[tex]\Delta T_b=i\times K_f\times m[/tex]
[tex]\Delta T_b=T_b-T_b^0=(116.3-114.1)^0C=2.2^0C[/tex] = elevation in boiling point
i= vant hoff factor = 1 (for non electrolyte)
[tex]K_b[/tex] = boiling point constant = [tex]9.43^0Ckg/mol[/tex]
m= molality
[tex]\Delta T_b=i\times K_b\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]
Weight of solvent (tin chloride)= 50.0 g =0.05 kg
Molar mass of unknown solute = M g/mol
Mass of unknown solute = 2.6 g
[tex]2.2=1\times 9.43\times \frac{2.6g}{M g/mol\times 0.05kg}[/tex]
[tex]M=222.8g/mol[/tex]
The possible formula for the compound would be [tex]InCl_3[/tex] as indium has valency of 3 and chlorine has valency of 1 has molecular mass almost equal to 222.8.
