1.

The freezing temperature of pure benzene is 5.40^(@)C. When 1.15g of naphthalene is dissolved in 100g of benzene, the resulting solution has a freezing point of 4.95^(@)C. The molal f.p. depression constant for benzene is 5.12, what is the molecular weight of naphthalene?

Answer»

Solution :MOLALITY `=(DletaT_(f))/(K_(f))=(5.40-4.95)/(5.12)=0.88` mole per `1000g`
`:.1000g` of the solvent contains `11.5g` of naphthalene
`:.` mol.wt.of naphthalene `=("weight in grams"//1000g)/("no.of moles"//1000g)` ………………..(RULE 1, Chapter 1)
`=(11.5)/(0.088)=130`


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