Please help with this problem, I really think my method is right, but the book has different ideas.

 

"The density of liquid nitrogen is 0.808 g/ml at -196 C. What volume of nitrogen gas at STP must be liquified to make 10.0 L of liquid nitrogen?"

 

so [math] \ce{N2(g) -> N2(l)} [/math]

 

The first step is to find the number of moles of N2, so

 

g = D*V so g = [imath] .808 $g/ml$ * 10.0L[/imath] = 8.08 g

 

moles = [imath] 8.08 g * \frac{1{mole}}{28.02g} [/imath] = .2883 moles

 

V = nRT/P = [math] \frac{.2883 moles* .08206L atm/Kmol * 298K}{1atm} [/math] = 7.05 L

 

the books answer is 6.46 * 10^3 L

 

What did I do wrong?

Don't forget that one Liter contains 1000 mL.

Man, I didn't know you could get that far in Chemistry and not even know your metric system... No offence or anything.

Man, I didn't know you could get that far in Chemistry and not even know your metric system... No offence or anything.

 

ouch...gotta say, that stung. Ok, so I change 10 L to 10,000 ml. That changes my answer to 7.05 L to 7.05 * 10^3 L...that's still not right. What else am I missing?

oops used wrong temp... never mind.

I dunno.

I dunno.

 

What do you mean?

 

I used room temp. (25 C) when I should have used standard temp. (0 C)

I think we must've posted at the same time, because when I replied, your second message wasn't there. Therefore my message corresponds to message #4, not message #5.

I think we must've posted at the same time, because when I replied, your second message wasn't there. Therefore my message corresponds to message #4, not message #5.

 

That's interesting... if you look at the time stamps you'll see that my message (5) was posted a full 22 hours before yours (6). Im pretty sure that's not at the same time.