What is the boiling point of a solution that contains 1.00 kg of water and a solute 3.00 moles of glucose ?

1 Answer
Junaid Mirza
Apr 18, 2018

#"101.536°C"#

Explanation:

When a solute is added in a solvent the boiling of solvent is elevated.

This elevation in boiling point (#Delta"T"_b#) is given by the formula

#Δ"T" = "K"_b"m"#

Where

  • #"K"_b =# Boiling point constant (#"0.512 °C/molal"# for water)
  • #"m ="# Molality of solution#= "Moles of solute"/"Mass of solvent (in kg)"#

Here solute is glucose & solvent is water.
Boiling point of pure water is #"100°C"#

#("T" - "100°C") = "0.512°C/molal" × "3.00 mol"/"1.00 kg"#

#"T = 100°C + 1.536°C = 101.536°C"#

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