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Chemistry 10 Online
OpenStudy (4815162342):

Which of the following aqueous solutions will have the lowest vapor pressure at 25 degrees Celsius? 1.0 M Fe(NO3)3 1.0 M C6H12O6 1.0 M KNO3 1.0 M CaCl2

OpenStudy (4815162342):

@matt6288

OpenStudy (matt6288):

A

OpenStudy (matt6288):

What did you think the answer was

OpenStudy (mww):

Vapour pressure is a good indicator of the volatility of the gaseous compound. If you have a high vp, it means it will more readily become gas at lower temperatures (i.e lower boiling points) compared to those with lower vps. So best to figure out the bps of each, or strength of bonding categorically. (BP is bit easier), I'd say sugar having no ionic bonds would have highest VP but check

OpenStudy (gollywolly):

That is probably way late but it's still fun to think about haha.

OpenStudy (gollywolly):

Something like this you would probably want to write down all the masses then determine the inter molecular forces. Like the sugar has hydrogen bonding which is a really strong inter molecular force. KNo_3 and Fe(No_3)3 can also hydrogen bond with water but not itself and they form ions I believe. CaCl2 is going to rely mostly on It's partial positive on on the Ca becuse Cl is pretty electronegative also it has the added weight weight of Ca all this makes it turn into ions in water. Also sterric hinderance plays a role in somethings ability to connect with other molecules

OpenStudy (gollywolly):

Molar mass of Fe(NO3)3 is 241.8597 g/mol; C6H12O6 is 180.1559 g/mol; KNO3 is 101.1032 g/mol; CaCl2 is 110.9840 g/mol. Fe(NO3)3 Is by far the heaviest so it will want to not become airborn plus NO3 can hydrogen bond; It will also produce 4 ions which will impede H2O's ability to escape; with this in mind I would say Fe(NO3)3.

OpenStudy (gollywolly):

ΔT = i Kb m. Substances which do not ionize in solution, like sugar, have i = 1. Substances which ionize into two ions, like NaCl, have i = 2. K_b is individual to each molecule. ex problem 11.4 g of ammonia in 200. g of water: 11.4 g / 17.031 g/mol = 0.6693676 mol 0.6693676 mol / 0.200 kg = 3.3468 m Determine bp elevation: Δt = i Kb m Δt = (1) (0.52 °C/m) (3.3468 m) Δt = 1.74 °C Sources http://www.chemteam.info/Solutions/BP-elevation.html

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