grams to liters

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grams to liters

Postby kp » Sat Nov 04, 2006 5:11 am

I need to convert 50g into liters. Is the following correct: 50g = 50cm^3 = 50 ml = .050L?
kp
 

Postby Dirtman » Sat Nov 04, 2006 10:52 am

The weight to volume ratio above is correct ONLY if using a substance that has a specific gravity of 1.0, such as water at 4┬║ C.

However, notwithstanding specific gravity, 50 cm^3 = 50 ml = 0.05 L
Dirtman
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Postby kp » Sat Nov 04, 2006 10:56 pm

Many thanks. I tried to look up specific gravity in my textbook (I'm in Freshman Inorganic chemistry) and it's not there so I guess I'm not supposed to know that yet.

What I'm trying to do is find the work done when 50 grams of tin dissolves in excess acid at 1 atm and 25 C. I know that work=-(P)(deltaV). In similar problems in the chapter, when we didn't know volume, we calculated it using V=nRT/P. However, in all those problems, the products & reactants were liquids and/or gasses with mass given in liters. For this problem, the initial volume is given in grams. I'm assuming I should use the same approach, i.e. calculate V-final and V-initial, determine delta V, and solve.

Would anyone be willing to comment on whether or not this is the right approach?
kp
 

conversions

Postby crackerjack » Tue May 08, 2007 10:14 am

you have no idea how long ive been looking for that conversion thanks
crackerjack
 

Postby obblman » Mon Aug 20, 2007 2:45 am

Since you have a pressure and volume, it's likely that they want you to convert your grams to moles (n) and use VP=nRT. Then take you liters (V) and apply to your equation. Because 50g does not equal 50 cm^3. Those are entirely separate. Mass and Volume. There is no conversion. (Unless you've got density)(density=g/L) Although it does beg the question how my teacher expects me to find grams of O2 from 2.54 liters without any other variables. :?
obblman
 


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