Okay so from following tha equation I got

v = (6.63 x 10^-34 J/s) / (2.36 x 10^-35 m/s) (.28349523125000003 kg)

v = 99.09591853


Does that sound right? Would the 99.0959 be in meters per second thus I should convert it to mph?

This post was edited by chicano on Jun 28 2015 06:17pm

don't slap the units on at the end. determine your units based on your calculations by carrying them forward. it will help you find problems if you did something wrong. eg if your calculations say your velocity is measured in kg

Not sure what else it could be? Is it wrong?

Yeah your units would be in m/s. will need to convert to mph
at this point you should write out your units and understand how to get different forms of the equations.

If you do that, you shouldn't have any question what units you get at the end. wrong or correct

Okay cool thank you. Here is another one from the same assignment:

The new vapor pressure of a solution of 3.25g of an unknown compound in 22.35 g of water is 114.9 torr at 55 degrees C. What is the molar mass of the unknown?

Made some conversions but not sure where to head next with this.

55 degrees C = 328.15 Kelvins
114.9 torr = .151 atmosphere
22.35g H20 = 1.241 moles
I'm also guessing 22.35g of water = 22.35 mls of water?

vapor pressure lowering

pressure_vapor,solution = X_h2o * P_vapor,h2o

you'll have to look up vapor pressure of h2o at 55c.

come up with an expression for the X_h2o, the molar fraction of h2o as a whole

No idea what you just said lol.


How about this one instead?

What mass percent of KBr in water would give you a -1.5O freezing point?

Freezing point depression,

Delta_Temperature = K_f * Molality_of_solution

K_f is the constant for freezing water

Molality_of_solution= mole sof solute/ 1kg of solution

But it doesn't give me grams or moles of the solute?

Is the KBr the solute and H2O the solution?

This post was edited by chicano on Jun 29 2015 06:51pm

KBr is the solute

yeah, it doesn't give you how many grams or moles of KBr
that's for you to calculate