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The freezing point of a solution contain...

The freezing point of a solution containing 2.40 g of a compound in 60.0g of benzene is `0.10^(@)"C"` lower than that of pure benzene. What is the molecular weight of the compound?`("K"_("f")" is " 5.12^(@)"C"//"m for benzene"`

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The correct Answer is:
`"2048 g/mol"`
Solute-B ,Solvent-A
`Delta"T"_("f")=("K"_("f")xx"w"_("B")xx1000)/("m"_("B")xx"w"_("A"))`
`0.10=(5.12xx2.40xx1000)/("m"_(B")xx60)`
`"m"_("B")=2048"gm mol"^(-1)`
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