I began by utilizing $T_f= K_f imes m $ (we"re given the $K_f$ that benzene is $5.12 space pu ^circ C/m$).
$4.20~ pu^circ C=(5.12 pu^circ C/m)(pum)$
molality = mole of solute every kg of solution
$implies 0.8203125$ molal= (unknown moles)/(kg solvent)
Am I claimed to usage kg of full solvent or kg the naphthalene? i can"t it seems ~ to number out whereby to go next.
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edited Mar 13 "18 in ~ 19:49
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molality = mole of solute per kg the solution
This is incorrect. Molality instead is mole of solute per kg that solvent. In her case, perform you know which among nepthalene or benzene is the solvent? Whichever the is, use its fixed in the formula"s denominator.
Moreover, your consumption of the formula, along the line "I began by using", is incorrect. The LHS need to be $Delta T_mathrm f~(=T_mathrm f2-T_mathrm f1)$ and also not $T_mathrm f$. The question is incomplete since it must have also mentioned the early freezing suggest of pure benzene solvent.
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answer Mar 14 "18 in ~ 2:16
Gaurang TandonGaurang Tandon
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