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Katyanochek1 [597]
3 years ago
5

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 = 9.43o

C 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?
Chemistry
1 answer:
Alexxx [7]3 years ago
4 0

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.

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