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Answer 1 · 2014-06-25T01:47:27

The effective molality of the solute is 4.9 mol·kg⁻¹.

We use the boiling point elevation expression

#ΔT_"b" = K_"b"m#

#ΔT_"b"# is the change in boiling point.

#K_"b"# is the molal boiling point elevation constant for the solvent.

#m# is the molality of the solute.

For water, #K_"b"# = 0.512 °C·kg·mol⁻¹

#ΔT_"b" = T_"b" – T_"b"^°# = 102.5 °C – 100.00 °C = 2.5 °C

#m = (ΔT_"b")/K_"b" = (2.5"°C")/(0.512"°C·kg·mol⁻¹")# = 4.9 mol·kg⁻¹

Hope this helps.