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प्रश्न
What is the freezing point of a liquid? The freezing point of pure benzene is 278.4 K. Calculate the freezing point of the solution when 2.0 g of a solute having molecular weight 100 g mol-1
is added to 100 g of benzene.
( Kf of benzene = 5.12 K kg mol-1 .)
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उत्तर
a) Freezing point: The freezing point of a liquid may be defined as the temperature at which the vapour pressure of solid is equal to the vapour pressure of liquid. A liquid freezes at a temperature at which the liquid and its solid coexist in equilibrium.
b) Solution:
Given:
Freezing point of pure solvent ( T° ) = 278.4 K
Mass of solute ( W2 ) = 2 g = 2 x 10-3 kg
Molar mass of solute ( M2 ) = 100 g mol-1 = 100 x 10-3 kg mol-1
Mass of solvent ( W1) = 100 g = 100 x 10-3 kg
Molal depression constant ( Kf )= 5.12 K kg mol-1.
To find: Freezing point of solution (T)
Formulae:
1. ΔTf = T° - T
2. ΔTf = `( "K"_"f" "W"_2 )/( "M"_2 "W"_1 )`
Calculation:
From formula (2),
ΔTf = `( "K"_"f" "W"_2 )/( "M"_2 "W"_1 )`
ΔTf = `( 5.12 xx 2 xx 10^-3 )/( 100 xx 10^-3 xx 100 xx 10^-3 )`
= `( 5.12 xx 2 )/10`
= 1.024 K
From formula (1),
ΔTf = T° - T
∴ T = T° - ΔTf
∴ T = 278.4 - 1.024 = 277.376 K
∴ Freezing point of the given benzene solution is 277.376 K.
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