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 $K_{f}$ 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}^{o} K g m o l^{- 1}

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Solution

The correct option is A $5.08 K K g m o l^{- 1}$
$Δ T = m K_{f}$
$Δ T = 3.55 - 5.45 = - 1.9$
GIven $0.374 M$ solution of tetrachloroethane
$K_{f} = \frac{Δ T}{M}$
$= \frac{1.9}{0.374}$
$= 5.08 K K g / m o l$

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Similar questions

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

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Q. When a

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Pure benzene freezes at 5.45oC at a certain but a 0.374 m solution of tetrachloroethane in thane benzene freezes at 3.55oC. The Kf for benzene is ?

The correct option is A 5.08KKgmol−1ΔT=mKfΔT=3.55−5.45=−1.9GIven 0.374M solution of tetrachloroethaneKf=ΔTM =1.90.374 =5.08KKg/mol

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 $K_{f}$ for benzene is ?

The correct option is A $5.08 K K g m o l^{- 1}$
$Δ T = m K_{f}$
$Δ T = 3.55 - 5.45 = - 1.9$
GIven $0.374 M$ solution of tetrachloroethane
$K_{f} = \frac{Δ T}{M}$
$= \frac{1.9}{0.374}$
$= 5.08 K K g / m o l$