Question

A 2.60 gram sample of a compound know to contain only indium and chlorine is dissolved in 50.0 g of tin(IV) chloride (Kb = 9.43oC kg mol-1). The normal boiling point is raised from 114.1oC for pure SnCl4 to 116.3oC for the solution. What is the molecular weight and probable molecular formula of the solute?

Answer 1

Answer: 1. The molecular weight of the compound is 222.8 g/mol

2. The probable molecular formula of the solute is $InCl_3$

Explanation:

Elevation in boiling point is given by:

$\Delta T_b=i\times K_f\times m$

$\Delta T_b=T_b-T_b^0=(116.3-114.1)^0C=2.2^0C$ = elevation in boiling point

i= vant hoff factor = 1 (for non electrolyte)

$K_b$ = boiling point constant = $9.43^0Ckg/mol$

m= molality

$\Delta T_b=i\times K_b\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

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

$2.2=1\times 9.43\times \frac{2.6g}{M g/mol\times 0.05kg}$

$M=222.8g/mol$

The possible formula for the compound would be $InCl_3$ as indium has valency of 3 and chlorine has valency of 1 has molecular mass almost equal to 222.8.