Pure benzece freezes at ${5.45}^{o} C$ at a certain place but a $0.374$ m solution of tetrachlororthane in benzene freezes at ${3.55}^{o} C$ .The $K_{f}$ for benzene is-

5.08 K K g m o l^{- 1}

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508 K K g m o l^{- 1}

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0.508 K K g m o l^{- 1}

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50.8 K K g m o l^{- 1}

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Solution

The correct option is D $5.08 K K g m o l^{- 1}$
given,
$T_{i n i t i a l} = {5.45}^{O} C$
$T_{f i n a l} = {3.55}^{O} C$
molality - $0.374 m$
Depression of freezing point $Δ T = T_{f i n a l} - T_{i n i t i a l}$
= $| 3.55 - 5.45 | = 1.9$
$Δ T = m K_{f}$
$K_{f}$ - freezing point constant
$K_{f} = \frac{Δ T}{m}$
$K_{f} = \frac{1.9}{0.374} = 5.08 K K g m o l^{- 1}$

Hence, option(a) is the correct answer.

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Similar questions

Q. Pure benzene freezes at

{5.45}^{o} C

at a certain but a

0.374

m solution of tetrachloroethane in thane benzene freezes at

{3.55}^{o} C

. The

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Pure benzene freezes at 5.45°C at a certain places but a 0.374 m solution of tetrachloroethane In benzene freezes at 3.55°C. The $K_{f}$ for benzene is

Q. Pure benzene freezes at

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Q. When a

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of tetrachloroethane is added to benzene, the solution freezes at

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]

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Pure benzece freezes at 5.45oC at a certain place but a 0.374 m solution of tetrachlororthane in benzene freezes at 3.55oC .The Kf for benzene is-

The correct option is D 5.08KKgmol−1given,Tinitial=5.45OC Tfinal=3.55OC molality - 0.374 m Depression of freezing point ΔT=Tfinal−Tinitial = |3.55−5.45|=1.9 ΔT=mKf Kf - freezing point constant Kf=ΔTm Kf=1.90.374=5.08 KKgmol−1 Hence, option(a) is the correct answer.

Pure benzece freezes at ${5.45}^{o} C$ at a certain place but a $0.374$ m solution of tetrachlororthane in benzene freezes at ${3.55}^{o} C$ .The $K_{f}$ for benzene is-