Question b1eda

Jun 30, 2017

${\chi}_{\text{NaOH}} = 0.159$

Explanation:

If I believe it to be true, you don't need the density value unless you're wanting to convert from molarity to mole fraction; we can calculate mole fraction from molality directly without any other information.

We're given that we have a $10.5 m$ $\text{NaOH}$ solution, which means there are $10.5$ $\text{mol NaOH}$ per $1$ $\text{kg H"_2"O}$:

$\left(10.5 \textcolor{w h i t e}{l} \text{mol NaOH")/(1color(white)(l)"kg H"_2"O}\right)$

All we have to do now is calculate the number of moles of water present, using its molar mass:

1cancel("kg H"_2"O")((10^3cancel("g H"_2"O"))/(1cancel("kg H"_2"O")))((1color(white)(l)"mol H"_2"O")/(18.02cancel("g H"_2"O"))) = color(red)(55.5 color(red)("mol H"_2"O"

The mole fraction of $\text{NaOH}$ is the ratio of moles of $\text{NaOH}$ to the total moles ($\text{NaOH"+"H"_2"O}$):

chi_ "NaOH" = (10.5cancel("mol NaOH"))/(10.5cancel("mol NaOH") + color(red)(55.5)cancel(color(red)("mol H"_2"O"))) = color(blue)(0.159

Jun 30, 2017

Here's what I got.

Explanation:

Assuming that the information you provided is correct and you're dealing with a $\underline{\text{10.5 molal}}$ solution of sodium hydroxide, then you don't really need to know its density in order to find the mole fraction of the solute.

As you know, molality is defined as the number of moles of solute present for every $\text{1 kg} = {10}^{3}$ $\text{g}$ of solvent.

In your case, a $\text{10.5-molal}$ solution of sodium hydroxide will contain $10.5$ moles of sodium hydroxide, the solute, per ${10}^{3}$ $\text{g}$ of water, the solvent.

Now, to make the calculations easier, let's pick a sample fo this solution that contains exactly ${10}^{3}$ $\text{g}$ of water. You already know from the molality of the solution that the sample contains $10.5$ moles of sodium hydroxide.

Use the molar mass of water to convert the mass of the solvent to moles

10^3 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "55.51 moles H"_2"O"

Now, the mole fraction of sodium hydroxide, ${\chi}_{\text{NaOH}}$, is equal to the number of moles of sodium hydroxide divided by the total number of moles present in the solution.

In your case, you will have

chi_ "NaOH" = (10.5 color(red)(cancel(color(black)("moles"))))/((10.5 + 55.51)color(red)(cancel(color(black)("moles")))) = color(darkgreen)(ul(color(black)(0.159)))#

The answer is rounded to three sig figs.

I highly recommend redoing the calculations using a different sample of solution, i.e. a solution that does not contain exactly ${10}^{3}$ $\text{g}$ of water $\to$ the mole fraction of the solute must be equal to $0.159$.

SIDE NOTE: If you're dealing with a $\text{10.5-M}$ solution, i.e. with molarity, not molality, then you must

• pick a $\text{1 L} = {10}^{3}$ $\text{mL}$ sample of this solution
• use its density to find the total mass of the solution
• use the molar mass of sodium hydroxide to find the mass of the solute present in the sample
• calculate the mass of water present in the sample
• use the molar mass of water to find the number of moles of water
• calculate the mole fraction of sodium hydroxide as shown above