1.

What is the weight of 1 mole of a solute, 0.132g of which in 29.7g of benzene, gave a freezing-point depression of 0.108^(@)C ? (K_(f) for benzene =5.12)

Answer»

SOLUTION :Molality `=(DeltaT_(f))/(K_(f))=(0.108)/(5.12)` ……….(EQN . 7)
and ALSO molality `=(0.32)/(M)XX(1000)/(29.7)m` (moles per `1000g`)
(`M=` mol.wt.or wt. of 1 mole of solute)
THUS,
`(0.132)/(M)xx(1000)/(29.7)=(0.108)/(5.12)`
`M=211.2g"mole"^(-1)`


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