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A chemist dissolves 1.3 mol NaCl in a 2.0-kg sample of water. What is the boiling point of this solution? (Assume that pure water boils at 100°C and that Kb for water = 0.512°C/m.)

Respuesta :

For this problem, the solution is exhibiting some colligative properties since the solute in the solution interferes with some of the properties of the solvent. We use equation for the boiling point elevation for this problem. We do as follows:

ΔT(boiling point)  = (Kb)mi
ΔT(boiling point)  = (0.512)(1.3/2.0)(2)
ΔT(boiling point)  = 0.67 degrees Celsius

T(boiling point) = 100 + 0.67 = 100.67 degrees Celsius

Answer:

100.67°C

Explanation:

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