anonymous 4 years ago 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?

1. anonymous

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.

2. anonymous

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.

3. anonymous

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)

4. anonymous

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?

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. anonymous

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