Question

an aqueous solution freezes at -1.5c calculate the boiling point of the solution (kf for water = 1.86 k kg mol-1 and kb for water = 0.51 k kg mol-1​

Answer 1

Answer:

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Answer 2

Given:

Depression constant in freezing point  $(K_{f})$  = $1.86 k$ $kg mol^{-1}$

Elevation constant in boiling point  $(K_{b} )$  = 0.51 $k$ $kg mol^{-1}$

Freezing point of solution $(T_{f} )$                 = - 1.5 °$C$

To Find:

Boing point of aqueous solution = ?

Solution:

Freezing point of solution $(T_{f} )$ is - 1.5 °$C$

Freezing point of solvent $(T_{f} )$  is  0 °$C$

Δ $T_{f}$   =  1.5 $K$

As we know from from depression in Freezing point equation

Δ $T_{f}$ = $K_{f}$ × $m$  

From elevation in B.P

Δ $T_{b}$ = $K_{b}$ × $m$  

We can now equate both equation as molality is same in both

Δ $T_{b}$ =  $K_{b}$  × Δ $T_{f}$ / $K_{f}$

Δ $T_{b}$ = 0.51 × 1.5 / 1.86

Δ $T_{b}$ = 0.41     (Elevation in B.P)

Boiling point of water = 373$K$

Boiling point of solution = 373 + 0.41 = 373.41$K$