Question

An aqueous solution of a non electrolyte solute boils at 100.52℃ the freezing point of the solution will be

Answer 1

The freezing point will be -0.52

Answer 2

The freezing point of the solution will be $-1.8^0</strong>C/tex]</h2><p><strong>Explanation:</strong></p><p><strong>Elevation in boiling point is given by:</strong></p><p>[tex]\Delta T_b=K_b\times m$

$\Delta T_b=T_b-T_b^0=(100.52-100)^0C=0.52^0C$ = elevation in boiling point

$K_b$ = boiling point constant = $0.52K/kgmol$

m= molality =?

$0.52^0C=0.52\times m$

$m=1.0$

Depression in freezing point is given by:

$\Delta T_f=K_f\times m$

$\Delta T_f=T_f^0-T_f=(0-(T_f)^0C$ = Depression in freezing point

$K_f$ = freezing point constant = $1.8K/kgmol$

m= molality = 1.0

$(0-(T_f)^0C=1.8\times 1.0$

$T_f=-1.8^0C$

Thus the freezing point of the solution will be $-1.8^0C$

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