# Question 242b5

Jul 17, 2016

Here's what I got.

#### Explanation:

You can't solve this problem without knowing the density of water. When the problem doesn't provide you with this information, you can assume that the density of water is

${\rho}_{\text{water" = "1 g cm}}^{- 3}$

This means that your sample of water has a mass of

1000 color(red)(cancel(color(black)("cm"^3))) * "1 g"/(1color(red)(cancel(color(black)("cm"^3)))) = "1000 g"

Now, use the molar masses of glucose, ${\text{C"_6"H"_12"O}}_{6}$, and water to calculate how many moles of each substance you have in your solution

100 color(red)(cancel(color(black)("g"))) * ("1 mole C"_6"H"_12"O"_6)/(180.156color(red)(cancel(color(black)("g")))) = "0.5551 moles C"_6"H"_12"O"_6

1000color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "55.51 moles H"_2"O"

The mole fraction of a component $i$ of a mixture, ${\chi}_{i}$, is calculated by dividing the number of moles of $i$ by the total number of moles present in the mixture

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {\chi}_{i} = \text{number of moles of i"/"total number of moles} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, the total number of moles present in solution will be

${n}_{\text{total" = n_"water" + n_"glucose}}$

${n}_{\text{total" = "0.5551 moles" + "55.51 moles" = "56.065 moles}}$

The mole fraction of glucose, your solute, will be

chi_"glucose" = (0.5551 color(red)(cancel(color(black)("moles"))))/(56.065color(red)(cancel(color(black)("moles")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(0.01)color(white)(a/a)|)))

The mole fraction of water, your solvent, will be

chi_"water" = (55.51color(red)(cancel(color(black)("moles"))))/(56.065color(red)(cancel(color(black)("moles")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(0.99)color(white)(a/a)|)))#

Notice that because your solution only contains glucose and water, you can find the mole fraction of water by

${\chi}_{\text{water" = 1 - chi_"glucose}}$

This once again gets you

${\chi}_{\text{water}} = 1 - 0.1 = 0.99$