What are the molal concentrations, and mole fractions, of a $385mg$ mass of $C_26H_46O$ dissolved in $40mL$ of chloroform?

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Answer 1 · 2017-06-28T19:21:24

$"Molality"$ $=$ $0.0249*mol*kg^-1$

$"Molality"$ $=$ $"Moles of solute"/"Kilograms of solvent"$

And here, $"molality"$ $=$ $((0.385*g)/(386.67*g*mol^-1))/(0.040*kg)=0.0249*mol*kg^-1$ .

On the other hand, the mole fraction, $chi$ , is dimensionless, and this is defined by the quotient.........

$chi_"solute"="Moles of solute"/"Total number of moles in solution"$

So here.........

$chi_"cholesterol"=("Moles of cholesterol")/("Moles of cholesterol+moles of chloroform")$

$=((0.385*g)/(386.67*g*mol^-1))/((0.385*g)/(386.67*g*mol^-1)+(40*g)/(119.38))=2.96xx10^-3$

And $"NECESSARILY"$ $chi_"chloroform"=1-chi_"cholesterol"=0.997$

Why $"necessarily?"$

What are the molal concentrations, and mole fractions, of a 385*mg385*mg mass of C_26H_46OC_26H_46O dissolved in 40*mL40*mL of chloroform?