My teacher has not explained how to find moles given pressure. I'm not sure how to start this problem. If anyone can help me at least get started, I would greatly appreciate it.
Question: At 25 degrees C, the vapor pressure of pure water is 23.76mmHg and that of a urea solution is 22.98mmHg. Calculate the molality of the solution.

I know that molality is (mols of solute)/(kg of solvent), but what relationship/formula will give me moles from this?

Question

My teacher has not explained how to find moles given pressure. I'm not sure how to start this problem. If anyone can help me at least get started, I would greatly appreciate it.
Question: At 25 degrees C, the vapor pressure of pure water is 23.76mmHg and that of a urea solution is 22.98mmHg. Calculate the molality of the solution.

I know that molality is (mols of solute)/(kg of solvent), but what relationship/formula will give me moles from this?

anonymous · Answer

I'll appreciate any help I can get. If you can at least point me in the right direction, I'll be happy with that too.

xishem · Answer

I think you might try approaching this using Raoult's Law?

$$P_A=X_AP_A^{\ \ \ 0}$$Where...
P_A is the vapor pressure of the solution in which A is the solvent.
X_A is the molefraction of solvent A.
P_A^0 is the vapore pressure of pure solvent A.

I'm a bit lost on this one too, but I'll try to work through it with you.

xishem · Answer

So, let's try Raoult's Law first...$$P_A=X_AP_A^{\ \ \ 0} \rightarrow X_A=\frac{P_A}{P_A^{\ \ \ 0}}$$$$X_A=\frac{22.98mmHg}{23.76mmHg}=0.9671_{717}$$\(sub-scripted numbers are just insignificant digits)

xishem · Answer

Now we know that the ratio of moles of solvent to moles of solute is...$$0.9671_{717}:0.0328_{2828}$$Now, let's try assuming we have 1 mole of solution...

In such a solution, there will be 0.9671717mol of water and 0.03282828mol of urea.

Given the definition of molality...$$b_i=\frac{n_i}{m_{solvent}}$$We need a mass of solvent...$$0.9672_{717}mol\ H_2O*\frac{18.015g\ H_2O}{1mol\ H_2O}=17.42_{3598}g\ H_2O$$Now that we have the mass of our solvent and mols of solute, we can calculate the molality...$$b_i=\frac{0.0328_{2828}\ mol\ urea}{17.42_{3598}\ g\ H_2O}$$$$b_i=\frac{0.00188_{4128}\ mol\ urea}{g\ H_2O}*\frac{1000g}{1kg}=1.88m$$Does this help?

anonymous · Answer

Thank you very much Xishem. You are a lifesaver :D

xishem · Answer

No problem (:. This was also a bit of review for me, so I relearned some concepts :D.

anonymous · Answer

haha for sure... this stuff is hard xD (for me at least)

anonymous · Answer

When you get finished with this one could you have a look at mine?