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`

Step-by-step explanation:

Elevation in boiling point is given by:

`\Delta T_b=i* K_f* 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* K_b* \frac{\text{mass of solute}}{\text{molar mass of solute}* \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* 9.43* (2.6g)/(M g/mol* 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.