Molality

[amcavoy]

At 25oC the vapor pressure of pure water is 23.76 mmHg and that of seawater is 22.98 mmHg. Assuming that seawater contains only NaCl, estimate its concentration in molality units.

X_1=\frac{n_1}{n_1+n_2}\implies n_2=\frac{n_1-X_1n_1}{X_1}

where n1 is the moles of solvent and n2 is the moles of solute.

22.98=X_1\left(23.76\right)\implies X_1=.9672

and 1000 g of water is equal to 55.49 mol (n1), so plugging this all in gives:

n_2=1.88\text{mol}

which would be the same as the molarity.

However, my textbook says that the molarity is .920 m. Where did I go wrong? Thanks a lot.

[Borek]

NaCl is dissolved.

Best,
Borek
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[amcavoy]

Right, Na++Cl-->NaCl.

Thanks for the help.

[bellanovela]

Can someone please explain this to me, because I'm not quite sure how to solve this problem. I only reached the part where I got the mole fraction of water and NaCl, but that's just it. I don't know what to do next. Help would be greatly appreciated. :(

[Borek]

You mean you have no idea how to convert molar fraction to molality?

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[bellanovela]

Okay, I found out how to convert it molality, but I keep getting 1.88 m not .920 m. How does the strong electrolyte/complete ionization of NaCl make a difference?

[Borek]

Check what Van't Hoff factor is.

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