# If 4.57*g of glucose are dissolved in 25.2*g of water, what are the mole fractions, chi_"each component"?

Jun 1, 2017

$\text{Mole fraction,}$ $\chi = \text{Moles of solute"/"Total moles of solution}$

#### Explanation:

And thus the sum of the mole fractions must be equal to one.......

${\chi}_{\text{glucose}} = \frac{\frac{4.75 \cdot g}{180.16 \cdot g \cdot m o {l}^{-} 1}}{\frac{4.75 \cdot g}{180.16 \cdot g \cdot m o {l}^{-} 1} + \frac{25.2 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1}} = 0.0185$

${\chi}_{\text{water}} = \frac{\frac{25.2 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1}}{\frac{4.75 \cdot g}{180.16 \cdot g \cdot m o {l}^{-} 1} + \frac{25.2 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1}} = 0.9815$

And these are dimensionless quantities............