# Mass of sugar that was added to the water (Raoult's law)

Water has a vapour pressure of $15.5~\mathrm{mmHg}$ at $25~\mathrm{^\circ C}$. An unknown amount of the non-volatile sugar glucose ($\ce{C6H12O6}$) is added to $1.31~\mathrm{kg}$ of water and the vapour pressure of the system is found to decrease by $0.13~\mathrm{mmHg}$. What mass of sugar, in grams, was added to the water?

Here it says that after the sugar was added to water, vapour pressure of the system is found to decrease by $0.13~\mathrm{mmHg}$.

Does it mean that before sugar was added, there was only one element $\ce{H2O}$?

And I used Raoult's law:

$$P_i = x_iP_i^*$$

where $P_i$ is the vapour pressure of the solvent, $x_i$ is the mole fraction of the solvent, and $P_i^*$ is the vapour pressure of the pure solvent.

So can I use this to find the partial pressure before the sugar was added?

I know that $1.31~\mathrm{kg}$ of water is $72.78~\mathrm{mol}$.

And if my assumption is right and before sugar was added there was only water, will the mole fraction before sugar was added just be $72.78$ since there was only $\ce{H2O}$?

• Note that the question contains a (not relevant) flaw: The vapour pressure of water at $T=25\ \mathrm{^\circ C}$ actually is $p=23.8\ \mathrm{mmHg}$, not $p=15.5\ \mathrm{mmHg}$. The value $p=15.5\ \mathrm{mmHg}$ corresponds to the vapour pressure of water at $T=18\ \mathrm{^\circ C}$.
– user7951
Nov 8 '15 at 13:59

Here it says that after the sugar was added to water, vapour pressure of the system is found to decrease by $0.13~\mathrm{mmHg}$.

Does it mean that before sugar was added, there was only one element $\ce{H2O}$?

One component, not element. Water isn't an element. But yes. The question does actually say it: "sugar is added to water".

I used Raoult's law: [...] So can I use this to find the partial pressure before the sugar was added?

Why do you want to find the partial pressure before it was added? It's already given to you. When there's only one component in the system, the partial pressure is the same as the total pressure, which is the vapour pressure of pure water: $15.5~\mathrm{mmHg}$.

In fact, the partial pressure after it was added is also given to you, it's just $(15.5 - 0.13)~\mathrm{mmHg}$. You already have all the pressure data you need. This question is giving you partial pressure data and asking you to find the mole fraction.

I know that $1.31~\mathrm{kg}$ of water is $72.78~\mathrm{mol}$.

Yes. Although if you use a more accurate molar mass for water, you will get a more accurate answer.

And if my assumption is right and before sugar was added there was only water, will the mole fraction before sugar was added just be $72.78$ since there was only $\ce{H2O}$?

No, the mole fraction would be $1$. The mole fraction of a component is the number of moles of that component, divided by the total number of moles.

The vapour pressure of pure water, which is $P_{\ce{H2O}}^*$ in Raoult's law, is $15.5~\mathrm{mmHg}$.

After you add a certain amount of sugar, the vapour pressure is decreased to $P_{\ce{H2O}} = 15.37~\mathrm{mmHg}$.

So, if you just plug these numbers into Raoult's law, you find:

\begin{align} x_{\ce{H2O}} &= \frac{P_{\ce{H2O}}}{P_{\ce{H2O}}^*} \\ &= \frac{15.37~\mathrm{mmHg}}{15.5~\mathrm{mmHg}} \\ &= 0.9916 \end{align}

$x_{\ce{H2O}}$ is the mole fraction of water after sugar is added. So:

\begin{align} x_{\ce{H2O}} &= \frac{n_{\ce{H2O}}}{n_\mathrm{total}} \\ 0.9916 &= \frac{n_{\ce{H2O}}}{n_{\ce{H2O}} + n_\mathrm{glucose}} \end{align}

The number of moles of water is:

$$\frac{1310~\mathrm{g}}{18.02~\mathrm{g~mol^{-1}}} = 72.70~\mathrm{mol}$$

This is what I said earlier about using a more accurate value of the molar mass. So:

$$0.9916 = \frac{72.70~\mathrm{mol}}{72.70~\mathrm{mol} + n_\mathrm{glucose}}$$

Rearranging:

$$n_\mathrm{glucose} = 0.6149~\mathrm{mol}$$

The molar mass of glucose is $180.16~\mathrm{g~mol^{-1}}$. So:

\begin{align} m_\mathrm{glucose} &= (0.6149~\mathrm{mol})(180.16~\mathrm{g~mol^{-1}}) \\ &= 110.8~\mathrm{g} \end{align}

• thank you very much ! the reason i could not solve this question, because i did not know that when there is only one component in the system, the partial pressure is the same as the total pressure. Nov 8 '15 at 17:29
• @student123 No problem. You might wish to read up more about vapour pressure, why it exists, how it is measured etc. - it is a confusing concept especially the first time you see it - it certainly was for me. Nov 8 '15 at 17:31