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When 20 g of naphthoic acid (C_(11)H_(8)O_(2)) is dissolved in 50 g of benzene (K_(f) = 1.72 "K kg mol"^(-1)), a freezing point depression of 2 K is observed. The van't Hoff factor (i) is

Answer»

`0.5`
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Solution :Molecular weight of napthoic acid, `C_(11)H_(8)O_(2)` is `172 g mol^(-1)`
The theoretical value of depression in freezing point = `k_(f)xx` MOLALITY `=1.72xx(20xx1000)/(172xx50)=4K`
`therefore` Van't Hoff factor i=
`("Observed value of colligative property")/("Theoretical value of colligative property")=(2)/(4)=0.5`


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