Pure benzene freezes at ${5.45}^{\circ} C$ . A $0.374 m$ solution of tetrachloroethane in benzene freezes at ${3.55}^{\circ} C$ . $K_{f}$ for benzene is:

0.508

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5.08

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50.8

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508

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Solution

The correct option is A $5.08$
$Δ T_{f} = K_{f} \times m$

$(5.45 - 3.55) = K_{f} \times 0.374$

$⟹ K_{f} = \frac{1.9}{0.374} = 5.08$

Hence, the correct option is $B$

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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

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

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{3.55}^{\circ} C

. Find the

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of benzene?
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{5.45}^{o} C

]

Q. Pure benzece freezes at

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at a certain place but a

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m solution of tetrachlororthane in benzene freezes at

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Pure benzene freezes at 5.45∘C. A 0.374 m solution of tetrachloroethane in benzene freezes at 3.55∘C. Kf for benzene is:

The correct option is A 5.08ΔTf=Kf×m(5.45−3.55)=Kf×0.374⟹Kf=1.90.374=5.08Hence, the correct option is B

Pure benzene freezes at ${5.45}^{\circ} C$ . A $0.374 m$ solution of tetrachloroethane in benzene freezes at ${3.55}^{\circ} C$ . $K_{f}$ for benzene is:

The correct option is A $5.08$
$Δ T_{f} = K_{f} \times m$

$(5.45 - 3.55) = K_{f} \times 0.374$

$⟹ K_{f} = \frac{1.9}{0.374} = 5.08$

Hence, the correct option is $B$