# Vapour pressure of a solute obtained during eudiometric analysis of methane combustion

### Problem

In a eudiometer $$\pu{14 ml}$$ of $$\ce{CH4}$$ and $$\pu{38.5 ml}$$ of $$\ce{O2}$$ was introduced. After explosion and cooling to initial temperature ($$\pu{23 °C}$$) the volume of eudiometer was found to be $$\pu{25 ml}$$. Calculate the vapour pressure(in $$\pu{mmHg}$$) of aqueous solution containing a non-volatile solute (A) at $$\pu{23 °C}$$. Mole fraction of A is $$0.1$$. The pressure of eudiometer is $$\pu{1 atm}$$ and is constant.

$$\pu{13.68 mmHg}$$

### Question

I wrote the balanced equation. Then applied Avogadro's law and found that the volume of gases in the eudiometer after reaction is $$\pu{24.5 ml}$$. So the extra $$\pu{0.5 ml}$$ must be due to some water which must have converted into gas. Using $$pV = nRT$$ I can calculate the amount of water evaporated, but I am unable to understand the last part of the question regarding solute A.

Is there some information missing in the question regarding solute A? I assumed it as the final solution after the reaction, but still don't know how to proceed.

• The vapour pressure of the solution is equal to the partial pressure of water in the gas phase. I'm not sure how this question was intended. You could get the vapour pressure of water at the given temperature from published data (pure water: 21.1 torr), or you could figure out the partial pressure of water in the solution. A complicating factor is that CO2 is soluble in water - maybe you are supposed to ignore that. If the non-volatile solute makes the solution acidic, not much CO2 will dissolve, but if it is basic, a lot will dissolve.
– Karsten
Commented Jun 11, 2019 at 17:36
• Could you edit your question to show your work that led to 24.5 mL of product? Are you including water in that, or is that just CO2?
– Karsten
Commented Jun 11, 2019 at 17:37
• "The pressure of the eudiometer is 1 atm and is constant." "After explosion..." :/
– Zhe
Commented Jun 11, 2019 at 19:40
• @Karsten Theis. The 24.5ml I have accounted is the volume for the excess O2+ volume of CO2 formed. Commented Jun 12, 2019 at 10:01
• 0.5 / 25•760*0.9=13.68, but this is not chemistry.
– Karsten
Commented Jun 12, 2019 at 10:52

I'll continue after volume of $$\ce H_2O$$ vapours = $$0.5 ml$$

Vapour pressure is equal to the partial pressure of water in gas phase.

$$P_{\ce H_2O} = P_{total}$$ × mole fraction of $$\ce H_2O$$

Where $$P_{total}$$ is $$1 atm$$ and mole fraction = $$0.5/25$$

So $$P_{\ce H_2O} = \frac{1}{50} atm = \frac{760}{50} mm Hg$$

Now for the solution of non volatile solute $$A$$

Using Raoult's law,

$$P_{solution}$$ = $$P_{\ce H2O}$$ × { 1 - (mole fraction of $$A$$) }

So $$P_{soln} = (760 × 0.9)/50 = 13.68 mm Hg$$