If the amount of a non-electrolyte dissolved is doubled but that of solvent is quadrupled, the elevation in boiling point of the solution will be

Question

If the amount of a non-electrolyte dissolved is doubled but that of solvent is quadrupled, the elevation in boiling point of the solution will be (A) doubled (B) halved (C) four times (D) unchanged

Tardigrade Admin · Accepted Answer

Δ Tb=(1000 Kb w2/w1 × M2) New Δ Tb'=(1000 Kb(2 w2)/(4 w1) × M2)=(1/2)((1000 Kb w2/w1 M2)) ∴ Δ Tb'=(Δ Tb/2)

Q. If the amount of a non-electrolyte dissolved is doubled but that of solvent is quadrupled, the elevation in boiling point of the solution will be

doubled

halved

four times

unchanged

$Δ T_{b} = \frac{1000 K _{b} w _{2}}{w _{1} \times M _{2}}$

New $Δ T_{b}^{'} = \frac{1000 K _{b} ( 2 w _{2} )}{( 4 w _{1} ) \times M _{2}} = \frac{1}{2} (\frac{1000 K _{b} w _{2}}{w _{1} M _{2}})$

$∴ Δ T_{b}^{'} = \frac{Δ T _{b}}{2}$

Q. If the amount of a non-electrolyte dissolved is doubled but that of solvent is quadrupled, the elevation in boiling point of the solution will be

doubled

halved

four times

unchanged

ΔTb​=w1​×M2​1000Kb​w2​​ New ΔTb′​=(4w1​)×M2​1000Kb​(2w2​)​=21​(w1​M2​1000Kb​w2​​) ∴ΔTb′​=2ΔTb​​

$Δ T_{b} = \frac{1000 K _{b} w _{2}}{w _{1} \times M _{2}}$

New $Δ T_{b}^{'} = \frac{1000 K _{b} ( 2 w _{2} )}{( 4 w _{1} ) \times M _{2}} = \frac{1}{2} (\frac{1000 K _{b} w _{2}}{w _{1} M _{2}})$

$∴ Δ T_{b}^{'} = \frac{Δ T _{b}}{2}$