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

Jun 28, 2017

$\text{Molality}$ $=$ $0.0249 \cdot m o l \cdot k {g}^{-} 1$

#### Explanation:

$\text{Molality}$ $=$ $\text{Moles of solute"/"Kilograms of solvent}$

And here, $\text{molality}$ $=$ $\frac{\frac{0.385 \cdot g}{386.67 \cdot g \cdot m o {l}^{-} 1}}{0.040 \cdot k g} = 0.0249 \cdot m o l \cdot k {g}^{-} 1$.

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

${\chi}_{\text{solute"="Moles of solute"/"Total number of moles in solution}}$

So here.........

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

$= \frac{\frac{0.385 \cdot g}{386.67 \cdot g \cdot m o {l}^{-} 1}}{\frac{0.385 \cdot g}{386.67 \cdot g \cdot m o {l}^{-} 1} + \frac{40 \cdot g}{119.38}} = 2.96 \times {10}^{-} 3$

And $\text{NECESSARILY}$ ${\chi}_{\text{chloroform"=1-chi_"cholesterol}} = 0.997$

Why $\text{necessarily?}$