(`K_(f)` for water `= 1.86 “K kg mol”^(-1)`)
A. `-0.372^(@)C`
B. `-0.520^(@)C`
C. `+0.372^(@)C`
D. `-0.570^(@)C`
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Correct Answer – D
Depression in freezing point
`DeltaT_(f)=K_(f)xxm`
Molality `=(W_(B)xx1000)/(M_(B)xxW_(A))=(68.5xx1000)/(342xx1000)`
`=(68.5)/(342)`m
`:. DeltaT_(f)=1.86xx(68.5)/(342)=0.372^(@)C`
`T_(f)=T_(f)^(@)-DeltaT_(f)=0.0-0.372=00.372^(@)C` .
Correct Answer – A
`Delta T_(f) = K_(f) m` Here, `m = 68.5/(342 xx 1)`
`:. Delta T_(f) = 0.2000 xx 1.86 = 0.3725^(@)C`
`T_(f) = 0^(@) -0.3725^(@)C= -0.3725^(@)C`
Correct Answer – A
`DeltaT=K_(f)xx(w_(B)xx1000)/(m_(B)xxw_(A))`
`=1.86xx(68.5xx1000)/(342xx1000)`
`=0.372`
Freezing point of solution `= – 0.372^(@)C`
`=- 0.372^(@)C`]