Question

An aqueous solution boils at 101°c what is the freezing point of the same solution

Answer 1

Answer:

the answer is - 4° c is your answer

Answer 2

Answer:  $-3.63^0C$

Explanation:

Elevation in boiling point :

$\Delta T_b=K_b\times m$

where,

$\Delta T_b=T_b-T_b^0$ =Elevation in boiling point

$K_b$ = boiling point constant of solvent= 0.512 °C/m

m = molality

$T_b$ = 101°C ,$T_b^0$ = 100°C

$\Delta T_b$= 101°C - 100°C = 1°C

$1^oC=0.512\times m$

$m=1.95$

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 = $1.86^0C/m$

m= molality

$0-T_f=1.86\times 1.95$

$0-T_f=3.63$

$T-f=-3.63^0C$

Thus freezing point of the same solution is $-3.63^0C$