Question

46. An aqueous solution freezes at -0.36°C. K and
K for water are 1.8 and 0.52 K kg mol-1
respectively then value of boiling point of solution at
1 atm pressure is
(1) 101.04°C
(2) 100.104°C
(3) 0.104°C
(4) 100°C
​

Answer 1

Answer: (2)   $100.104^0C$

Explanation:

Depression in freezing point is given by:

$\Delta T_f=K_f\times m$

$\Delta T_f=T_f^0-T_f=(0-(-0.360))^0C=0.36^0C$ = Depression in freezing point

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

m= molality

$0.36=1.8\times m$

$m=0.2mol/kg$

Now elevation in boiling point is given by:

$\Delta T_b=K_b\times m$

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

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

m= molality  = 0.2 mol/kg

$(T_b-100)^0C=0.52\times 0.2=0.104$

$T_b=(0.104+100)=100.104^0C$

Thus value of boiling point of solution at  1 atm pressure is $100.104^0C$

Answer 2

Explanation:

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